$\Psi: V_1\longrightarrow V_2$ such that $\Psi(F)=\Phi \circ F$. Let $V_1=Lin(\mathbb{R_{<=2}[x]},\mathbb{R_{<=2}[x]})$ and $V_2=Lin(\mathbb{R_{<=2}[x]},\mathbb{R})$ two vector spaces and let 
$$\Phi \in V_2, \qquad \Phi(p(x))=p'(1).$$
Consider 
$$\Psi: V_1\longrightarrow V_2\qquad \Psi(F)=\Phi \circ F.$$
How can I find the dimension of the Kernel of $\Psi$ and a basis?
Then, consider $W=\{F\in V_1 : F(x − 1) = 0\}$. What's the dimension of $W$?
 A: Consider the base $\mathcal B =\{x^2,x,1\}$ of $\mathbb R_{\leq 2}[x]$ and consider the coordinates isomorphism:
$$
\mathbb R_{\leq 2}[x] \rightarrow \mathbb R^3,\qquad a_2x^2+a_1x+a_0\mapsto (a_2,a_1,a_0)^T
$$
By this isomorphism an element of $\text{Lin}(\mathbb R_{\leq 2}[x],\mathbb R_{\leq 2}[x])$ is just a $3\times 3$ matrix, and an element of $\text{Lin}(\mathbb R_{\leq 2}[x],\mathbb R)$ is a $1\times 3$ matrix.
In particular how does appear the element $\Phi$ read in coordinate? We have $\Phi(a_2x^2+a_1x+a_0) = 2a_2 + a_1$, so it is represented by the matrix:
$$
\mathcal M(\Phi) = \left(\begin{matrix}
2 & 1 & 0 
\end{matrix}
\right)
$$
And how does appear the function $\Psi$? We have $\Psi(F)=\Phi\circ F$; since $F$ is represented by a $3\times 3$ matrix we have:
$$
\Psi\left(
F
\right) = \left(\begin{matrix}
2 & 1 & 0 
\end{matrix}
\right) \circ \left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right)=
\left(
\begin{matrix}
2a_{1,1} + a_{2,1} & 2a_{1,2} + a_{2,2} & 2a_{1,3} + a_{2,3}
\end{matrix}
\right)
$$
Now it's easy to find $\ker\Psi$ and a base of it: 
\begin{gather}
\ker\Psi=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right) \in V_1 \ \text{such that} \ \ \ \begin{matrix} 2a_{1,1} + a_{2,1}=0,\\ 2a_{1,2} + a_{2,2}=0,\\ 2a_{1,3} +a_{2,3}=0
\end{matrix}
\right\}\\
=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
-2a_{1,1} & -2a_{1,2} & -2a_{1,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right)\in V_1\right\}
\end{gather} 
So $\ker\Psi$ has dimesion $6$. An explicit base is really easy to see now. 

To find a base of $W$ again consider coordinates. The polynomial $x-1$ is represented by the vector $(0,1,-1)^T$, so:
$$
\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{matrix}
\right) \left(\begin{matrix}
0\\
1\\
-1
\end{matrix}\right) = \left(\begin{matrix}
0\\
0\\
0
\end{matrix}\right) 
$$
And with the same proof as above you will find 
$$
W=\left\{\left(
\begin{matrix}
a_{1,1} & a_{1,2} & a_{1,2}\\
a_{2,1} & a_{2,2} & a_{2,2}\\
a_{3,1} & a_{3,2} & a_{3,2}
\end{matrix}
\right)\in V_1
\right\}
$$
So also $W$ has dimension equal to $6$.

I tried to do this exercise without coordinates, but I did not find a simple way. To see the dimensions of the subspaces you can think to how many conditions are the properties of the subspace. For example in the second part, $F(x-1)=0$ gives you $3$ conditions and the dimensions is $9-3=6$. This is not always true because sometimes some conditions are equivalent, but you can use this "fact" in order to have an idea of the result.
