I agree with your representation of $\displaystyle T$. Now, $\displaystyle e_{i}$ is a standard notation for basis vectors. I would guess that
$\displaystyle e_{1}=\left[\begin{matrix}1\\ 0\end{matrix}\right]$, and
$\displaystyle e_{2}=\left[\begin{matrix}0\\ 1\end{matrix}\right]$.
So, what do you suppose $\displaystyle T(e_{1})$ is?
Neither. And, no, the matrix form is not $\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$.
You are told that the linear transfromation is "from $\displaystyle R^2$ to $\displaystyle R^3$" so you will be multiplying a matrix by a "column matrix" with 2 entries (actually 1 column, 2 rows as a matrix) so there must be two columns in you matrix to be able to multiply. The result is in [itex]R^3[/itex] so it will be a "column matrix" with 3 entries (1 column, 3 rows) which means your matrix must have 3 rows to give 3 results. The matrix representing this linear transformation must be "2 by 3" (2 columns, 3 rows).
Do as Ackbeet suggested: since you titled this "Standard matrices and basis vectors" he, and I, assume that $\displaystyle e_1= \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $\displaystyle e_2= \begin{bmatrix}0 \\ 1\end{bmatrix}$.
You are told that $\displaystyle T(x_1, y_1)= (3x_1- x_2, -x_1+ x_2, 5x_2)$ so what is T(1, 0)? What is T(0, 1)?
Now use the fact that $\displaystyle \begin{bmatrix}a & b\\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ c \\ e\end{bmatrix}$ and $\displaystyle \begin{bmatrix}a & b\\ c & d \\ e & f\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix}b \\ d \\ f\end{bmatrix}$ to find a, b, c, d, e, and f.