Answer
4 and 1.
Work Step by Step
The singular values of the matrix are the square roots of the eigenvalues of $A^TA$.
$A^TA=\begin{bmatrix}
2&0\\3&2
\end{bmatrix}\begin{bmatrix}
2&3\\0&2
\end{bmatrix}=\begin{bmatrix}
4&6\\6&13
\end{bmatrix}$. The characteristic equation is $16-17\lambda+\lambda^2$ and its eigenvalues are 16 and 1, so the singular values are 4 and 1.