I am not sure on how to go about this. Please provide clear explanations.
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The question as phrased isn’t really meaningful, since you didn’t specify a relation on the empty set. However, there is only one, so I’ll assume that it’s the one that you meant. A relation on a set $A$ is a subset of $A\times A$. Since $\varnothing\times\varnothing=\varnothing$, the only subset of $\varnothing\times\varnothing$ is $\varnothing$. Thus, the question can be interpreted as:
Let’s recall the definitions
The thing to notice here is that each of these is a universally quantified statement: something must be true of every element, pair of elements, or trio of elements of $A$. Thus, in order to show that $R$ is not reflexive, you must find an $a\in A$ such that $a\,\not R\,a$; in order to show that $R$ is not antisymmetic, you must find elements $a,b\in A$ such that $a\,R\,b$ and $b\,R\,a$, but $a\ne b$; in order to show that $R$ is not transitive, you must find elements $a,b,c\in A$ such that $a\,R\,b,b\,R\,c$, and $a\,\not R\,c$; and in order to show that $R$ is not total, you must find elements $a,b\in A$ such that $\,\not R\,b$ and $b\,\not R\,a$. The crucial point is that in each case you must find elements of $A$ that actually have certain properties. If such elements don’t exist, then $R$ is reflexive (or antisymmetric, transitive, or total).
In your question $A$ is $\varnothing$, the empty set. Is it possible to find elements of $\varnothing$ that have certain properties? Is it possible to find elements of $\varnothing$ at all?
If a set is totally ordered, then it is already partially ordered.
It is meaningless to say whether any set $A$ is partially or totally ordered. You need to give a set $A$ and a partial ordering $\prec$ on $A$ for this question to make sense.
However, there does exists a total ordering on the $\emptyset$. It is the empty relation. More concretely, you can also think of the $\in$ relation restricted to $\emptyset$.
The above answers confirm that $\emptyset$ with the empty ordering is in fact a linear order. I would just like to add that it is in fact also well-ordered under this ordering since it is linearly ordered and every non-empty subset of $\emptyset$ has a least element. This is true vacuously since there are no nonempty subsets at all.