Answer: \(p=4\), \(q=2\).

Write all terms in the first equation as powers of \(2\):

\(8^{p+1}=2^{3(p+1)}, \qquad 4^q=2^{2q}.\)

So

\(\frac{2^{3(p+1)}}{2^{2q}}=2^{11}.\)

Equating powers gives

\(3p+3-2q=11.\)

Thus

\(3p-2q=8. \tag{1}\)

For the second equation, write everything as powers of \(3\):

\(27^{1/3}=3, \qquad 9^{3q}=3^{6q}.\)

So

\(\frac{3^{2p+5}}{3}=3^{6q}.\)

Equating powers gives

\(2p+5-1=6q,\)

so

\(2p-6q=-4. \tag{2}\)

Solving (1) and (2), from (2)

\(p=3q-2.\)

Substitute into (1):

\(3(3q-2)-2q=8.\)

Hence

\(7q=14, \qquad q=2.\)

Then

\(p=3(2)-2=4.\)

Therefore

\(\boxed{p=4,\quad q=2}.\)