I'll be assuming your matrices are real, but nothing much would change if they were complex.
The first thing to note is that the the kernel is a subset of the domain and the image is a subset of the codomain - this trips a lot of people up. To check if a matrix is in the kernel, you apply $F$ to it (like you correctly suggest for matrix $b$). However to check if a matrix $A$ is an "image", you need to see if there exists a matrix $M$ such that $F(M)=A$. So matrix $a$ is not an image, because every matrix in the image of $F$ has zero in the upper right and lower left. Matrix $c$ is an image, since
$$
F\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=c
$$
To find bases for the kernel and image, you can start with a base for $M(2\times 2)$. The standard base is the four matrices:
$$
e_1=\left(\begin{array}{cc}1 & 0\\ 0& 0\end{array}\right)\quad e_2=\left(\begin{array}{cc}0 & 1\\ 0& 0\end{array}\right)\quad e_3=\left(\begin{array}{cc}0 & 0\\ 1& 0\end{array}\right)\quad e_4=\left(\begin{array}{cc}0 & 0\\ 0& 1\end{array}\right)
$$
Then, you know that any matrix $A\in M(2\times 2)$ can be written as $$
A=ae_1+be_2+ce_3+de_4. $$ Now apply $F$ to your basis elements:
$$
F(e_1)=e_1\quad F(e_2)=0\quad F(e_3)=0\quad F(e_4)=e_4
$$
Since $F$ is linear, we can describe its range by taking an arbitrary element of $M(2\times 2)$, writing in terms of our base $\{e_1,e_2,e_3,e_4\}$, and applying $F$ to it:
$$
F(ae_1+be_2+ce_3+de_4)=aF(e_1)+bF(e_2)+cF(e_3)+dF(e_4)=ae_1+de_4
$$
Thus the image can be described using the two basis matrices $e_1$ and $e_4$:
$$
\text{Im}F=\{\alpha e_1+\beta e_4:\alpha,\beta\in\Bbb{R}\}
$$
Like you suggest, the kernel can be described as any matrix with zero diagonal. Now that we have a basis for $M(2\times 2)$, we can write this succinctly as
$$
\text{Ker}F=\{\alpha e_2+\beta e_3:\alpha,\beta\in\Bbb{R}\}
$$
You should check to make sure these are subspaces.
Bonus question: show that any matrix $A\in M(2\times 2)$ can be written as the sum of a matrix in the kernel of $F$ and a matrix in the image of $F$.
Notice that
$$T(a,b,c)=a(1+x^2)+b(-1+x)+c(-x-x^2)\\=(a-c)(1+x^2)+(b-c)(-1+x)$$
so
$$T(a,b,c)\in \operatorname{span}(1+x^2,-1+x)$$
and by the rank-nullity theorem we see that the rank of $T$ is $2$ so $(1+x^2,-1+x)$ is a basis for the image of $T$.
Best Answer
For the kernel you must find a basis for the space of vector $\vec{x}$ satisfying $$A\vec{x}=\vec{0}.$$
The image is all $\vec{y} \in \mathbb{R}[x]_3$ such that $A\vec{x}=\vec{y}$. So we are trying to find a basis for the space $$A\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$ By matrix multiplication, this space is given by the linear combination of columns of $A$: $$\vec{C}_1x_1+\vec{C}_2x_2+\vec{C}_3x_3+\vec{C}_4x_4,$$ where $C_n$ is the $n^{th}$ column of $A$.
Answer for Kernel:
Answer for Column Space: