Input to $T$ the vectors of the basis $B$ and coordinatize the outputs relative to $C$,
\begin{align*}
\vectrep{C}{\lt{T}{1}}&=
\vectrep{C}{
\begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}
}
=
\vectrep{C}{
2\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+
1\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+
1\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
}
=
\colvector{2\\1\\1}\\
\vectrep{C}{\lt{T}{1+x}}&=
\vectrep{C}{
\begin{bmatrix}1 & 4 \\ 4 & 1\end{bmatrix}
}
=
\vectrep{C}{
1\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+
4\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+
1\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
}
=
\colvector{1\\4\\1}\\
\vectrep{C}{\lt{T}{1+x+x^2}}&=
\vectrep{C}{
\begin{bmatrix}2 & 3 \\ 3 & 0\end{bmatrix}
}
=
\vectrep{C}{
2\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+
3\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+
0\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
}
=
\colvector{2\\3\\0}
\end{align*}
Applying Definition MR we have the matrix representation
\begin{equation*}
\matrixrep{T}{B}{C}=
\begin{bmatrix}2 & 1 & 2 \\ 1 & 4 & 3 \\ 1 & 1 & 0 \end{bmatrix}
\end{equation*}
To compute $\lt{T}{3+5x-2x^2}$ employ Theorem FTMR,
\begin{align*}
\lt{T}{3+5x-2x^2}&=
\vectrepinv{C}{\matrixrep{T}{B}{C}\vectrep{B}{3+5x-2x^2}}\\
&=\vectrepinv{C}{\matrixrep{T}{B}{C}\vectrep{B}{(-2)(1)+7(1+x)+(-2)(1+x+x^2)}}\\
&=\vectrepinv{C}{
\begin{bmatrix}2&1&2\\1&4&3\\1&1&0\end{bmatrix}
\colvector{-2\\7\\-2}
}\\
&=\vectrepinv{C}{\colvector{-1\\20\\5}}\\
&=
(-1)\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+
20\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+
5\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\\
&=
\begin{bmatrix}-1 & 20 \\ 20 & 5\end{bmatrix}
\end{align*}
You can, of course, check your answer by evaluating $\lt{T}{3+5x-2x^2}$ directly.