Then $\star \star$ As shown in Figure a, the base of the quadrilateral pyramid $P-ABCD$ is a square, $PA \perp$ the base $ABCD$, $PA=AD=2$, points $M$ and $N$ are on edges $PD$ and $PC$ respectively, and $PC \perp$ plane $AMN$.
(1) Prove that $AM \perp PD$; (2) Find the size of the dihedral angle $P-AM-N$; (3) Find the angle between line $CD$ and plane $AMN$. Figure $\mathbf{a}$
(1) Since quadrilateral $ABCD$ is a square, therefore, $CD \perp AD$. Also, since $PA \perp$ the base $ABCD$, then $PA \perp CD$. Hence, $CD \perp$ plane $PAD$. Since $AM \subset$ plane $PAD$, then $CD \perp AM$. Given $PC \perp$ plane $AMN$, we have $PC \perp AM$. Therefore, $AM \perp$ plane $PCD$. Hence, $AM \perp PD$.
(2) Since $AM \perp$ plane $PCD$ (already proven), then $AM \perp PM, AM \perp NM$. Therefore, $\angle PMN$ is the plane angle of the dihedral angle $P-AM-N$. Also, since $PN \perp$ plane $AMN$, then $PN \perp NM$. In the right triangle $\triangle PCD$, $CD=2, PD=2\sqrt{2}$, then $PC=2\sqrt{3}$.
Since $PA=AD, AM \perp PD$, then $M$ is the midpoint of $PD$. Therefore, $PM=\frac{1}{2}PD=\sqrt{2}$. By $\mathrm{Rt} \triangle PMN \sim \mathrm{Rt} \triangle PCD$, we get $MN=\frac{CD \cdot PM}{PC}$. Then $\cos \angle PMN=\frac{MN}{PM}=\frac{CD}{PC}=\frac{2}{2\sqrt{3}}=\frac{\sqrt{3}}{3}$. Hence, $\angle PMN=\arccos \frac{\sqrt{3}}{3}$. Therefore, the dihedral angle $P-AM-N$ is $\arccos \frac{\sqrt{3}}{3}$.
(3) As shown in Figure b, extend $NM, CD$ to intersect at point $E$. Since $PC \perp$ plane $AMN$, then $NE$ is the projection of $CE$ in plane $AMN$. Therefore, $\angle CEN$ is the angle between $CD$ (i.e., $CE$) and plane $AMN$. Also, since $CD \perp PD, EN \perp PN$, then $\angle CEN=\angle MPN$. In the right triangle $\triangle PMN$, since $\sin \angle MPN=\frac{MN}{PM}=\frac{\sqrt{3}}{3}$, and $\angle MPN \in \left(0, \frac{\pi}{2}\right)$, then $\angle MPN=\arcsin \frac{\sqrt{3}}{3}$. Therefore, the angle between $CD$ and plane $AMN$ is $\arcsin \frac{\sqrt{3}}{3}$.