\(\bf \begin{cases}
(n^4)^p=n^{12}\to &n^{4\cdot p}=n^{12}\to 4p=12
\\
&\qquad \uparrow \\
&\textit{same bases, thus}\\
&\qquad \downarrow
\\
n^3\cdot n^q=n^6\to &n^{3+q}=n^6\to 3+q=6
\end{cases}
\\\\\\
thus\implies
\begin{cases}
4p=12\implies p=\frac{12}{4}
\\\\
3+q=6\implies q=6-3
\end{cases}
\\\\
\\\\
then\implies p\cdot q=\boxed{?}\)
