13. (13 points) As shown in Figure 2, 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 $AF \perp PB$. Find:
(1) the length of $PA$;
(2) the sine of the dihedral angle $B-AF-D$.


13. (1) As shown in Figure 3, connect $B D$, intersecting $A C$ at point $O$. Taking $O$ as the origin, the directions of $\overrightarrow{O B}$, $\overrightarrow{O C}$, and $\overrightarrow{A P}$ are the positive directions of the $x$-axis, $y$-axis, and $z$-axis, respectively, to establish the spatial rectangular coordinate system $O-x y z$.
$\Rightarrow z=2 \sqrt{3}$ (negative value discarded).
Thus, $|\overrightarrow{P A}|=2 \sqrt{3}$, which means the length of $P A$ is $2 \sqrt{3}$.
(2) From (1), we know
$$
\begin{array}{l}
\overrightarrow{A D}=(-\sqrt{3}, 3,0), \overrightarrow{A B}=(\sqrt{3}, 3,0), \\
\overrightarrow{A F}=(0,2, \sqrt{3}) .
\end{array}
$$

Let the normal vector of plane $F A D$ be $\boldsymbol{n}_{1}=\left(x_{1}, y_{1}, z_{1}\right)$, and the normal vector of plane $F A B$ be $\boldsymbol{n}_{2}=\left(x_{2}, y_{2}, z_{2}\right)$.
$$
\begin{array}{l}
\text { By } \boldsymbol{n}_{1} \cdot \overrightarrow{A D}=0, n_{1} \cdot \overrightarrow{A F}=0 \\
\Rightarrow\left\{\begin{array}{l}
-\sqrt{3} x_{1}+3 y_{1}=0, \\
2 y_{1}+\sqrt{3} z_{1}=0
\end{array}\right. \\
\Rightarrow \boldsymbol{n}_{1}=(3, \sqrt{3},-2) \text {. } \\
\Rightarrow\left\{\begin{array}{l}
\sqrt{3} x_{2}+3 y_{2}=0, \\
2 y_{2}+\sqrt{3} z_{2}=0
\end{array}\right. \\
\Rightarrow \boldsymbol{n}_{2}=(3,-\sqrt{3}, 2) \text {. } \\
\end{array}
$$

By $\boldsymbol{n}_{2} \cdot \overrightarrow{A B}=0, n_{2} \cdot \overrightarrow{A F}=0$

Thus, the cosine value of the angle between the normal vectors $\boldsymbol{n}_{1}$ and $\boldsymbol{n}_{2}$ is
$$
\cos \left\langle\boldsymbol{n}_{1}, \boldsymbol{n}_{2}\right\rangle=\frac{\boldsymbol{n}_{1} \cdot \boldsymbol{n}_{2}}{\left|\boldsymbol{n}_{1}\right|\left|\boldsymbol{n}_{2}\right|}=\frac{1}{8} .
$$

Therefore, the sine value of the dihedral angle $B-A F-D$ is $\frac{3 \sqrt{7}}{8}$. Since $B C=C D$, $\triangle B C D$ is an isosceles triangle. Also, $A C$ bisects $\angle B C D$, so $A C \perp B D$.
Thus, $O C=C D \cos \frac{\pi}{3}=1$.
Given $A C=4 \Rightarrow A O=A C-O C=3$.
Also, $O D=C D \sin \frac{\pi}{3}=\sqrt{3}$, hence,
$$
\begin{array}{l}
A(0,-3,0), B(\sqrt{3}, 0,0), \\
C(0,1,0), D(-\sqrt{3}, 0,0) .
\end{array}
$$

Since $P A \perp$ the base $A B C D$, and $F$ is the midpoint of $P C$,
$$
\begin{array}{l}
\text { Let } P(0,-3, z) \text {, so, } F\left(0,-1, \frac{z}{2}\right) \text {. } \\
\quad \text { By } A F \perp P B \Rightarrow \overrightarrow{A F} \cdot \overrightarrow{P B}=0 \\
\quad \Rightarrow\left(0,2, \frac{z}{2}\right) \cdot(\sqrt{3}, 3,-z)=0 \\
\quad \Rightarrow 6-\frac{z^{2}}{2}=0
\end{array}
$$