My attempt:
Since $(trace(A), 5*trace(a), - trace(A))$ is only 0 if the trace of the matrix is 0 and the space of the 2x2 matrices with trace 0 is 3-dimensional, so by the rank-nullity theorem the image is 1-dimensional.
At this point how do I get a basis for the image and the kernel?
First choose bases for the set of 2x2 matrices $\mathcal{M}_2(\mathbb{R})$ and $\mathbb{R}^3$.
Let $\epsilon_1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\epsilon_2 = \begin{pmatrix}0&1\\0&0\end{pmatrix}$, $\epsilon_3 = \begin{pmatrix}0&0\\1&0\end{pmatrix}$, $\epsilon_4 = \begin{pmatrix}0&0\\0&1\end{pmatrix}$, then $\mathcal{E} = \{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4 \} $ is a basis for $\mathcal{M}_2(\mathbb{R})$. For $\mathbb{R}^3$, choose your standard basis $E = \{e_1,e_2,e_3,e_4\}$.
Now we proceed to find the matrix of $T$ with respect to $\mathcal{E}$ on the left and $E$ on the right. We obtain:
$$M(T)_{\mathcal{E}}^{E}=\begin{pmatrix}1&0&0&1\\5&0&0&5\\-1&0&0&-1\end{pmatrix}$$
Now put it in Row-Echelon form to get your basis for your kernel. For the image, you can very quickly notice that the columns of $ M(T)_{\mathcal{E}}^{E}$ span $Im(T)$ and the only linearly independent subset of the columns is $\{(1, 5, -1)\}$, so there's your basis.