20. As shown in Figure 1, given that the base of the pyramid $P-ABCD$ is a square with side length 4, and $PD \perp$ the base $ABCD$. Let $PD=6$, and $M$, $N$ be the midpoints of $PB$ and $AB$ respectively.
(1) Find the volume of the tetrahedron $P-DMN$;
(2) Find the tangent value of the dihedral angle $M-DN-C$.
20. As shown in Figure 2, let $A C$ and $B D$ intersect at $O$, and connect $M O$ and $P N$.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
M O \perp \text { base } A B C D, \\
M O=\frac{1}{2} P D=3 .
\end{array}\right. \\
\end{array}
$$
(1) Since $N$ is the midpoint of $A B$, $\therefore S_{\triangle O V B}=\frac{1}{2} S_{\triangle X X B}=4$.
$$
\begin{array}{l}
\therefore V_{P-\text { UMV }}=V_{P-D N B}-V_{N-U V E} \\
=\frac{1}{3} S_{\triangle \triangle N B} \cdot P D-\frac{1}{3} S_{\triangle A V H} \cdot M O \\
=\frac{1}{3} S_{\triangle O V E}(P D-M O)=\frac{1}{3} \cdot 4 \cdot(6-3)=4 \text {. } \\
\end{array}
$$
(2) Draw $O K \perp D N$ through $O$, and connect $K M$. Then, by the theorem of three perpendiculars, $M K \perp D N$.
Thus, $\angle M K O$ is the plane angle of the dihedral angle $M-D N-C$.
Connect $O N$, and it is easy to find that $S_{\triangle O O V V}=\frac{1}{2} S_{\triangle O V B}=2$.
Also, $D N=\sqrt{A D^{2}+A N^{2}}=2 \sqrt{5}$,
$$
O K=\frac{2 S_{\triangle \text { OUV }}}{D N}=\frac{2}{\sqrt{5}} \text {. }
$$

Therefore, $\tan \angle M K O=\frac{M O}{O K}=\frac{3 \sqrt{5}}{2}$ is the desired result.