What is the difference between zero scalar and zero vector? Zero vector has zero value in the given vector space. So, it is different from zero scalar. Zero vector is additive identity of the given vector space whereas zero scalar is not. I do understand this academic distinction. But I still have following $2$ doubts.
(i) In one dimensional space, is zero scalar same as to zero vector?
According to me - NO. In $1D$ space, the vectors may be expressed in real number notation (instead of matrix notation) where absolute value of real number indicates magnitude and sign indicated direction. Then, we are using 'real number notation' to represent not a real number, but a vector in $1D$. What we represent is not real number. So, a zero vector in $1D$ space is indeed expressed as $0$ in 'real number notation'. However it is not representing the real number zero (which is member of set of real numbers) but representing zero vector (which is a member of vectors in 1D space). So, in $1D$ space, a zero vector may be represented by number $0$. But it is not the real number $0$. 
Analogy -: A directed segment represents a vector. It is not a vector.
(ii) Is zero speed same as zero velocity? In general, if we define a scalar quantity $Q_1$ which is magnitude of a vector quantity $Q_2$, then is zero $Q_1$ same as zero $Q_2$?
According to me - NO. When speed is zero, then the velocity is zero and vice-versa. However, zero speed is not equal to zero velocity.
 A: For question 1) the answer is "usually no". For example, let $n$ be a positive integer greater than $1$ and let $V$ be a one dimensional subspace of $\mathbb R^n$. The zero vector in $V$ is certainly not the scalar $0$. The reason I say "usually" no is that if you view $\mathbb R$ as a vector space over $\mathbb R$, then the zero vector happens to be equal to the zero scalar. You could cook up some other examples like that.
For question 2, speed is a scalar and velocity is a vector. If an object has speed $0$, then its velocity is the zero vector, but its speed is not equal to its velocity. (They could not be equal because they are not even the same type of mathematical object.)
To be more concrete, let's say that I introduce a coordinate system in my lab and measure that my speed (in meters/sec) is the number $0$.  Then my velocity (in meters / sec) is $(0,0,0)$.  And $0 \neq (0,0,0)$.  
Here's another way to make the same point.  Suppose that a particle's position at time $t$ is $f(t)$, where $f:\mathbb (a,b) \to \mathbb R^3$ is a differentiable function.  The particle's velocity at time $t_0$ is $f'(t_0)$, and the particle's speed at time $t_0$ is $\| f'(t_0)\|$.  Suppose that the particle's speed at time $t_0$ is the number $0$.  Then the particle's velocity at time $t_0$ is $f'(t_0) = (0,0,0)$.  And again, $0 \neq (0,0,0)$.
A: Perhaps it would be useful to look at the differences between vectors and scalars instead of focusing exclusively on the zero vector and the zero of a field.
The set of vectors in a vector space forms an Abelian group under addition. This means that there is an identity element, namely the zero vector, and every vector $\mathbf{v}\in V$ has an additive inverse, $-\mathbf{v}$, and vectors are associative under addition. Some simple examples of Albelian groups are $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, and $(\mathbb{Z}/n\mathbb{Z},+)$. There are also multiplicative Albelian groups like $(\mathbb{Q},\times)$, $(\{z\in\mathbb{C}:|z|=1\},\times)$, and $(\{a^n:n\in\mathbb{Z}\},\times)$. In these types of groups, it more useful to think of the identity as 1 instead of 0.
Fields are also Abelian groups under addition, and this means that they too have a zero element that behaves precisely the same way as a zero vector. This may be where confusion arises. However, fields also have an additional operation that vectors do not and that is multiplication. Any two elements of a field can be multiplied together to give another element of the field. Elements of a field can also be divided. Some common examples of fields are $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$.
There is also the condition that the set of vectors $V$ must be compatible with the base field $\mathbb{F}$, so certain combinations of groups and fields will not form vector spaces. For example, if we take take our set of vectors to be $V=(\mathbb{R}^n,+)$ and the base field to be $\mathbb{C}$, will we have a vector space? No, because for any vector $\mathbb{v}\in V$, the vector $i\mathbb{v}$ is not an element of $V$ and closure under scalar multiplication is necessary for the pair $(V, \mathbb{F})$ to form a vector space. 
It can be easy to confuse the zero vector with the zero of a field, but they are not the same in any dimension. So your answer to the first question is correct. In my intro course on linear algebra, they denoted the zero element of the field by writing a little $\mathbb{F}$ under the zero like this: $0_\mathbb{F}$ to help us distinguish it from the zero vector which we denoted with a little $V$ underneath the zero like this $0_V$. 
With regards to your last question, zero speed is indeed the same as zero velocity, and this is true of any scalar quantity defined as the norm of a vector since $||\mathbf{v} ||=0_\mathbb{F}\iff \mathbf{v} =0_V$. They are distinct entities, but one implies the other.
A: *

*Strictly, no, but isomorphically, yes. Any finite-dimensional space $V$ is isomorphic to $K^n$, where $K$ is the associated scalar field and $n=\dim(V)$. If $\{v_1,\ldots,v_n\}$ is a base, you can write any $v\in V$ uniquely as $v=\sum_{j=1}^n k_jv_j$; the isomorphism is then $v\leftrightarrow (k_1,\ldots k_n)$. In particular, in a one-dimensional space, $\vec0$ and $0$ are equal under that isomorphism since  $\vec0=0\cdot v_1$.

*Very strictly speaking, the two are not equal but equivalent. If $Q_1=||Q_2||$ then
$$Q_1=0\Leftrightarrow Q_2=0,$$
if $||\cdot||$ is any norm (magnitude). This, however, is axiomatic.

A: Look up the definition of a vector space from an algebraic point of view. Briefly, for a vector space you need a group $(G,+_G)$ and a field $(k,\cdot_k,+_k)$ combined in an intrinsic way.
In $G$ since it is a group you can combine elements by the group operation and in $k$ you have two operations with their respective rules (commutativity, associativity and distributivity). And the connection between these two structures is given by
$$
\lambda\cdot_k(g+_Gh)=\lambda\cdot_kg+\lambda\cdot_kh\qquad \forall \lambda\in k,\quad g,h\in G
$$
and
$$
g\cdot_k(\lambda+_k\mu)=g\cdot_k\lambda+_Gg\cdot_k\mu\quad \forall g\in G\quad \lambda,\mu\in k
$$
Try not to think about dimensions in a concrete case (like matrices or vectors) think like we have a sack of elements where we can combine elements according to group rules. And we have an another sack of things of a different kind where we can combine the elements according to field rules. Both structures need to have  a zero element to work but these zero elements can be of totally different kind/sort. I think exactly this is what disturbs you since in the concrete case you are looking at they happened to have the same kind but still coming from the different sacks in the abstract setting. So try to think about the zero scalar as coming from $k$ and the zero vector coming from $G$. If you only need the linear algebraic setting it may seem a bit of an overshoot to look at the abstract setting but I am sureit will clarify things if you have the time to sit down with it for some time.
A: The scalar  $0$ is the zero from the field  $F $ that the vector space is a vector space over. ..  This is different from the $n$-dimensional zero vector ${(0,\dots, 0)}$,  where  $n $ is the dimension of the vector space...(in the  $1$-dimensional case there is only one zero;  but the concepts are still different :  on the one hand the $1$-tuple $(0)$, as opposed to simply the element $0\in F$, on the other. ..  The context is different.)  Also,  vectors can only be added...  whereas scalars multiply vectors to give new vectors. ..
Secondly, speed is a scalar,  the magnitude of the velocity,  which is a vector and thus has a direction in addition to a length. ..
For example,  in uniform circular motion the speed is constant and the velocity vector stays perpendicular to the acceleration, which is towards the center  of the circle...  This is instructive,  and applicable, in that, the velocity vector keeps changing  (in direction ), whilst the speed is constant. ...
A: You do not want to know about "academic" (would you mean "algebraic"?) arguments. So, proceeding pragmatically here are my answers to your two questions.
(i) A vector, or vector quantity, is a quantity whose complete determination is given by a magnitude and a direction, with the only exception when the magnitude is zero. In this case the direction does not make sense, and the vector is only its magnitude: there are no two different vectors with magnitude zero and different directions. This happens independently of the dimension.
When you measure or compute a vector quantity you need to measure or compute its magnitude and its direction, but if its magnitude results to be zero you have done, no direction should be determined.
So the zero vector quantity is the zero scalar quantity (in whatever dimension).
(ii) Same arguments of (i) applies here with a notable difference. A velocity vector is a bound vector, while in (i) vectors were free vectors. So here a bound vector is the triple made up of a magnitude, a direction, and an application point. So here you cannot say that the zero velocity (vector quantity) is the zero speed (scalar quantity), but you need to use the concept of a scalar in a point (a map) and now can only say that:
the zero speed in a given point is the zero velocity in that same point (in whatever dimension).
Please only note that using algebraic arguments involving the standard algebraic definition of vector, scalar, vector space, field, equality (instead of those given here), answers to (i) and (ii) are neither positive nor negative, because scalars and vectors, and so the zero scalar and the zero vector in any dimension, are not even comparable in the general setting of abstract algebra, where only structures are studied, that is, how operations work on abstract entities, independently of the real nature of the entities: so you can have a given real entity that can play the role of a scalar and a vector at the same time, but abstract algebra alone can tell them apart, only their distinct roles. Then some context can be added thus passing from abstract algebra to applied algebra, and in so doing an equality relation can be defined on the set of real entities which will involve characteristics not belonging to the abstract concept of vector, scalar, number, ... but belonging to the real nature of the entities. Only now yow can start asking about equality of a vector and a scalar. I can give you details if you need but it's a bit more involved.
