13. As shown in the figure, in the quadrilateral pyramid $P-ABCD$, $PA \perp$ base $ABCD$, $BC=CD=2$, $AC=4$, $\angle ACB=\angle ACD=\frac{\pi}{3}$, $F$ is the midpoint of $PC$, and $AF \perp PB$.
(1) Find the length of $PA$;
(2) Find the sine of the dihedral angle $B-AF-D$.
(1) Connect $B D$, intersecting $A C$ at point $M$, connect $P M$, intersecting $A F$ at point $G$. From $B C=2, A C=4, \angle A C B=\frac{\pi}{3}$ $\Rightarrow A B=2 \sqrt{3}$, similarly $A D=2 \sqrt{3}$. Also, $B C=C D=2, \angle B C D=\frac{2 \pi}{3} \Rightarrow B D=2 \sqrt{3}$, thus $\triangle A B D$ is an equilateral triangle $\Rightarrow A M=3, M C=1$.
$B D \perp A C, B D \perp P A \Rightarrow B D \perp$ plane $P A C \Rightarrow B D \perp A F$,
also $A F \perp P B \Rightarrow A F \perp$ plane $P B D \Rightarrow A G \perp P M$. Therefore, in $\triangle P A C$, let $P A=x$,
$P C=\sqrt{16+x^{2}} \Rightarrow A F=F C=\frac{\sqrt{16+x^{2}}}{2} \Rightarrow \cos \angle F A C=\frac{4}{\sqrt{16+x^{2}}}$,
and $\angle F A C=\angle A P M, \cos \angle A P M=\frac{x}{\sqrt{9+x^{2}}} \Rightarrow \frac{4}{\sqrt{16+x^{2}}}=\frac{x}{\sqrt{9+x^{2}}}$
$\Rightarrow 16\left(9+x^{2}\right)=x^{2}\left(16+x^{2}\right) \Rightarrow x=2 \sqrt{3}$.
(2) From (1), we know $\cos \angle F A C=\frac{2}{\sqrt{7}} \Rightarrow A G=\frac{6}{\sqrt{7}} \Rightarrow G B=G D=\sqrt{12-\frac{36}{7}}=\frac{4 \sqrt{3}}{7}$,

In $\triangle G B D$, $\cos \angle B G D=\frac{\frac{48}{7}+\frac{48}{7}-12}{2 \cdot \frac{48}{7}}=\frac{1}{8}$, and from (1) we know $\angle B G D$ is the plane angle of the dihedral angle $B-A F-D$, so its sine value is $\frac{3 \sqrt{7}}{8}$.