Answer: \(2q+3p=13\), \(2p+pq=12\), \(p=3\), \(q=2\).

The determinant of \(M\) is

\(\det M=2q-p(-3)=2q+3p.\)

Since \(\det M=13\),

\(2q+3p=13.\)

Now compute the top-right entry of \(M^2\):

\(\begin{pmatrix}2&p\\-3&q\end{pmatrix}\begin{pmatrix}2&p\\-3&q\end{pmatrix}.\)

The top-right entry is

\(2p+pq.\)

It is given that this entry is \(12\), so

\(2p+pq=12.\)

From \(2q+3p=13\),

\(q=\dfrac{13-3p}{2}.\)

Substitute into \(2p+pq=12\):

\(2p+p\left(\dfrac{13-3p}{2}\right)=12.\)

Multiply by \(2\):

\(4p+13p-3p^2=24.\)

So

\(3p^2-17p+24=0.\)

Factor:

\((3p-8)(p-3)=0.\)

Since \(p\) is an integer,

\(p=3.\)

Then

\(2q+3(3)=13,\)

so

\(2q=4,\)

and hence

\(\boxed{p=3,\quad q=2}.\)