Answer
9 and 1.
Work Step by Step
The singular values of the matrix are the square roots of the eigenvalues of $A^TA$.
$A^TA=\begin{bmatrix}
3&8\\0&3
\end{bmatrix}\begin{bmatrix}
3&0\\8&3
\end{bmatrix}=\begin{bmatrix}
73&24\\24&9
\end{bmatrix}$. The characteristic equation is $81-82\lambda+\lambda^2$ and its eigenvalues are 81 and 1, so the singular values are 9 and 1.
