13. (13 points) As shown in Figure 2, in the 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 $O-x y z$.
Since $B C = C D$, we know that $\triangle B C D$ 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 $P C$, let $P(0, -3, z)$, so $F\left(0, -1, \frac{z}{2}\right)$.
$$
\begin{array}{l}
\text{By } A F \perp P B \Rightarrow \overrightarrow{A F} \cdot \overrightarrow{P B} = 0 \\
\Rightarrow \left(0, 2, \frac{z}{2}\right) \cdot (\sqrt{3}, 3, -z) = 0 \\
\Rightarrow 6 - \frac{z^2}{2} = 0
\end{array}
$$
$\Rightarrow z = 2 \sqrt{3}$ (negative value discarded).
Thus, $|\overrightarrow{P A}| = 2 \sqrt{3}$, meaning 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 $\boldsymbol{n}_{1} = (x_{1}, y_{1}, z_{1})$ and the normal vector of plane $F A B$ be $\boldsymbol{n}_{2} = (x_{2}, y_{2}, z_{2})$.
$$
\begin{array}{l}
\text{By } \boldsymbol{n}_{1} \cdot \overrightarrow{A D} = 0, \boldsymbol{n}_{1} \cdot \overrightarrow{A F} = 0 \\
\Rightarrow \left\{\begin{array}{l}
-\sqrt{3} x_{1} + 3 y_{1} = 0, \\
2 y_{1} + \sqrt{3} z_{1} = 0
\end{array}\right. \\
\Rightarrow \boldsymbol{n}_{1} = (3, \sqrt{3}, -2). \\
\text{By } \boldsymbol{n}_{2} \cdot \overrightarrow{A B} = 0, \boldsymbol{n}_{2} \cdot \overrightarrow{A F} = 0 \\
\Rightarrow \left\{\begin{array}{l}
\sqrt{3} x_{2} + 3 y_{2} = 0, \\
2 y_{2} + \sqrt{3} z_{2} = 0
\end{array}\right. \\
\Rightarrow \boldsymbol{n}_{2} = (3, -\sqrt{3}, 2).
\end{array}
$$

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

Therefore, the sine of the dihedral angle $B - A F - D$ is $\frac{3 \sqrt{7}}{8}$.