20. As shown in the figure, the base of the pyramid $P-ABCD$ is a square with side length 4, and $PD \perp$ the base $ABCD$. Given $PD=6$, and $M, N$ are the midpoints of $PB, AB$ respectively.
(1) Find the volume of the tetrahedron $P-DMN$;
(2) Find the tangent value of the dihedral angle $M-DN-C$.


20. Let $A O, B D$ intersect at $O$, connect $M O, P N$.
$\left.\begin{array}{l}\because M O / / P D \\ P D \perp \text { base } A B C D\end{array}\right\} \Rightarrow\left\{\begin{array}{l}M O \perp \text { base } A B C D, \\ M O=\frac{1}{2} P D=3 .\end{array}\right.$
(I) $\because N$ is the midpoint of $A B$,
$$
\begin{aligned}
\therefore S_{\triangle D N B} & =\frac{1}{2} S_{\triangle D A B}=4 . \\
\therefore V_{P-D M N} & =V_{P-D N B}-V_{M-D N B} \\
& =\frac{1}{3} S_{\triangle D N B} \cdot P D-\frac{1}{3} S_{\triangle D N B} \cdot M O \\
& =\frac{1}{3} S_{\triangle D N B}(P D-M O) \\
& =\frac{1}{3} \cdot 4 \cdot(6-3)=4 .
\end{aligned}
$$
(II) Draw $O K \perp D N$ through $O$, connect $K M$, then by the theorem of three perpendiculars, $M K \perp D N$. Therefore, $\angle M K O$ is the plane angle of the dihedral angle $M-D N-C$.
Connect $O N$, it is easy to find that $S_{\triangle C D N}=\frac{1}{2} S_{\triangle D N B}=2$.
Also, $D N=\sqrt{A D^{2}+A N^{2}}=2 \sqrt{5}, O K=\frac{2 S_{\triangle O D N}}{D N}=\frac{2}{\sqrt{5}}$. Thus, $\tan \angle M O K=\frac{M O}{O K}=\frac{3 \sqrt{5}}{2}$ is the required value.