Let $A$ be the matrix of $T:P_2\to P_2$ with respect to basis $B=\{v_1,v_2,v_3\}$. Find $T(v_1)$
$$A=\begin{bmatrix} 1 & 3 & -1 \\ 2 & 0 & 5 \\ 6 & -2 & 4 \end{bmatrix}$$
$v_1=3x+3x^2$
$v_2=-1+3x+2x^2$
$v_3=3+7x+2x^2$
First part of question asks to find $T(v_1)_B$.
$$T(v_1)_B=[T]_{P_2}^{B}\cdot (v_1)_{P_2}=\begin{bmatrix} 1 & 3 & -1 \\ 2 & 0 & 5 \\ 6 & -2 & 4 \end{bmatrix}\cdot\begin{bmatrix} 0 \\ 3 \\ 3 \end{bmatrix}=\begin{bmatrix} 6 \\ 15 \\ 6 \end{bmatrix}$$
Where $v_1$ is our vector, $B$ is the basis, $P_2=\{1,x,x^2\}$, $A=[T]_{P_2}^{B}$
Assuming I did this part of the question correct, now the next part asks be to find $T(v_1)$, with respect to no basis.
I think this makes sense: since we already found $T(v_1)_B$, that means that the vector $[6,15,6]$ can be written as a linear combinations of the vectors in the basis $B$. Hence we have:
$$6+15x+6x^2=r_1(3x+3x^2)+r_2(-1+3x+2x^2)+r_3(3+7x+2x^2)$$
Cleaning up a bit we get:
$$6+15x+6x^2=(-r_2+3r_3)1+(3r_1+3r_2+7r_3)x+(3r_1+2r_2+2r_3)x^2$$
Giving us the solution set:
$$\color{red}{-r_2+3r_3=5}, \color{green}{3r_1+3r_2+7r_3=15}, \color{blue}{3r_1+2r_2+2r_3=6}$$
Solving for $r_1,r_2,r_3$, we should have $T(v_1)$. Is this correct?
You did the first part incorrectly. It looks like where you tripped up is that $A=[T]_B^B$, not $[T]_{P_2}^B$. $[v_1]_B=[1,0,0]$, so $[Tv_1]_B$ is just the first column of $A$ (and similarly for $v_2$ and $v_3$. This is a handy property to remember: in general, the columns of the matrix of a linear transformation are the images of the basis vectors, expressed relative to the “output” basis.
That aside, you also computed the basis conversion in the wrong direction. If you have $[w]_B=[a,b,c]$, this means that $w=av_1+bv_2+cv_3$, so to recover the polynomial that corresponds to $[Tv_1]_B$, you simply need to compute this linear combination.