You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $A\in\mathbb{R}^{(2,2)}$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_{2,1}=0$. Thus
$$
N(T)=\{A\in\mathbb{R}^{(2,2)}\mid a_{2,1}=0\}=\{\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}\in\mathbb{R}^{(2,2)}\}
$$
The classical basis for $\mathbb{R}^{(2,2)}$ is the basis associated with the canonical basis for $\mathbb{R}^4$:
$$
\begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\1 &0\end{pmatrix}, \begin{pmatrix}0 &0\\0 &1\end{pmatrix}
$$
You may check that these are indeed linear independent and span $\mathbb{R}^{(2,2)}$. Coming back to the null space of $T$, you can see that every matrix $A\in N(T)$, i.e. every matrix of the form
$$
\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}
$$
can be created as a linear combination of
$$
\begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\0 &1\end{pmatrix}
$$
with
$$
\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}=a_{1,1}\begin{pmatrix}1 &0\\0 &0\end{pmatrix} + a_{1,2} \begin{pmatrix}0 &1\\0 &0\end{pmatrix}+ a_{2,2}\begin{pmatrix}0 &0\\0 &1\end{pmatrix}
$$
as the defining condition for $N(T)$ is a zero at position $a_{2,1}$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $\mathrm{dim}(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=\mathrm{dim}(\mathbb{R}^{(2,2)})=\mathrm{dim}(N(T))+\mathrm{dim}(R(T))=3+\mathrm{dim}(R(T))
$$
Therefore, $\mathrm{dim}(R(T))=1$, i.e. as $T:\mathbb{R}^{(2,2)}\to\mathbb{R}$, we have $R(T)=\mathbb{R}$. A typical basis for $\mathbb{R}$ as a vector space is the number $1$.
Best Answer
An arbitrary element of $W$ is $$ A=\begin{pmatrix} a & a \\a & b\end{pmatrix}=a\begin{pmatrix} 1 & 1 \\1 & 0\end{pmatrix}+b\begin{pmatrix} 0 & 0 \\0 & 1\end{pmatrix} $$ So a basis of $W$ is $\{\begin{pmatrix} 1 & 1 \\1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\0 & 1\end{pmatrix}\}$