Dual spaces and linear functionals

We have to solve the following:

Let $V$ be an $n$-dimensional vector space over $\mathbb{K}$ and let $B=(v_{1}, ... ,v_{n})$ be the basis of $V$. Every $x \in V$ can be described as $$x=\sum_{k=1}^{n}c_{k}v_{k}$$ with $c_{k} \in \mathbb{K}$.

1. Let $j \in \mathbb{N}: 1\leq j\leq n$. Show that the map $w_{j}: V\rightarrow \mathbb{K}$, which transforms $x$ into $c_j$, is a linear functional!
2. Show that $(w_{1}, ... ,w_n)$ is a basis of the dual space $V^*$.
3. Show that the linear map $V\rightarrow V^*$ is an isomorphism.

What I did:

1. I don't understand the first question at all...
2. I would say that it's the basis of $V^*$ because $\dim(V)=\dim(V^*)$, meaning the basis must have $n$ elements.
3. I showed that it's bijective, is that enough?

I'm really stuck with this problem, it would be very nice if someone could help!

1) For prove this you just have to apply the definition of a linear functional for every $w_j$, i.e $\forall$ x, y $\in V$ $w_j(x + y)=w_j(x)+w_j(y)$ and $\forall$ $\alpha \in K$ $w_j(\alpha x)=\alpha w_j(x)$

I) Let $x=\sum x_i v_i$ and $x=\sum y_i v_i$ elements of $V$ and $j \in \{1, ... , n\}$, then

$w_j(x+y)=w_j(\sum x_i v_i + \sum y_i v_i) = w_j(\sum (x_i + y_i) v_i)=x_j + y_j = w_j(x)+w_j(y)$

II) Let $\alpha \in K$ and $x$ as before

$w_j(\alpha x)=w_j(\alpha \sum x_i v_i)=w_j(\sum \alpha x_i v_i)=\alpha x_j = \alpha w_j(x)$

2) For this we just have to show that $\{w_1, ... , w_n\}$ generate $V$ (the part of linear independence is free from the dimension as you say in your comments), so let $f \in V^*$ a linear functional and $x \in V$, then

$f(x)=f(\sum x_iv_i)=\sum x_if(v_i)=\sum w_j(x)f(v_i)$ i.e $f$ is a linear combination of $\{w_1, ... , w_n\}$ (the first time that you see this could be weird, but give it some time and you will got it!).

3) Define the function $\phi:V \rightarrow V^*$ such that $\sum x_i v_i \mapsto \sum x_iw_i$ (the good part of this function is that send basis into basis, so you can use this for prove the isomorphism easily).

1. Each vector $x$ in $V$, once you chose a basis, can be represented as a $n$-uple $(c_1,\dots,c_n) \in \mathbb{K}^n$. You can consider the map: $w_j: V \to \mathbb{K}$ defined by $w_j(v) = c_j$, i.e. the map that gives you the $j$-th coordinate. You have to show this is linear.

2. In order to show that $\{w_1,\dots,w_n\}$ is a basis you have to show that each $f \in V^*$ can be written as $\sum_{j=1}^n \alpha_j w_j$

3. You have to show that the map $V \to V^*$ is bijective, linear and also the inverse is linear