13. Given quadrilateral $A B C D$ satisfies $A D \| B C, B A=A D=D C=\frac{1}{2} B C=a, E$ is the midpoint of $B C$, and $\triangle B A E$ is folded along $A E$ to form $\triangle B_{1} A E$, such that plane $B_{1} A E \perp$ plane $A E C D$, and $F$ is the midpoint of $B_{1} D$.
(1) Find the volume of the pyramid $B_{1}-A E C D$;
(2) Find the sine of the angle between plane $A D B_{1}$ and plane $E C B_{1}$.
(1) Take the midpoint $M$ of $A E$, and connect $B_{1} M$. Since $B A=A D=D C=\frac{1}{2} B C=a$, and $\triangle A B E$ is an equilateral triangle, then $B_{1} M=\frac{\sqrt{3}}{2} a$. Also, because the plane $B_{1} A E \perp$ plane $A E C D$, it follows that $B_{1} M \perp$ plane $A E C D$, so
$$
V=\frac{1}{3} \times \frac{\sqrt{3}}{2} a \times a \times a \times \sin \frac{\pi}{3}=\frac{a^{3}}{4}
$$
(2) Connect $M D$, and take $M E, M D, M B_{1}$ as the $x, y, z$ axes, respectively, then
$$
\begin{array}{c}
E\left(\frac{a}{2}, 0,0\right), C\left(a, \frac{\sqrt{3}}{2} a, 0\right), A\left(-\frac{a}{2}, 0,0\right), D\left(0, \frac{\sqrt{3}}{2} a, 0\right), B_{1}\left(0,0, \frac{\sqrt{3}}{2} a\right) \\
\overrightarrow{E C}=\left(\frac{a}{2}, \frac{\sqrt{3} a}{2}, 0\right), \overrightarrow{E B_{1}}=\left(-\frac{a}{2}, 0, \frac{\sqrt{3} a}{2}\right), \overrightarrow{A D}=\left(\frac{a}{2}, \frac{\sqrt{3} a}{2}, 0\right), \overrightarrow{A B_{1}}=\left(\frac{a}{2}, 0, \frac{\sqrt{3} a}{2}\right)
\end{array}
$$

Let the normal vector of plane $E C B_{1}$ be $\vec{v}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right),\left\{\begin{array}{l}\frac{a}{2} x^{\prime}+\frac{\sqrt{3}}{2} a y^{\prime}=0 \\ -\frac{a}{2} x^{\prime}+\frac{\sqrt{3}}{2} a z^{\prime}=0\end{array}\right.$, let $x^{\prime}=1$, then $\vec{v}=\left(1,-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$;
Let the normal vector of plane $A D B_{1}$ be $\vec{u}=(x, y, z),\left\{\begin{array}{l}\frac{a}{2} x+\frac{\sqrt{3}}{2} a y=0 \\ \frac{a}{2} x+\frac{\sqrt{3}}{2} a z=0\end{array}\right.$, let $x=1$, then $\vec{u}=\left(1,-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}\right)$, then
$$
\cos =\frac{1+\frac{1}{3}-\frac{1}{3}}{\sqrt{1+\frac{1}{3}+\frac{1}{3}} \times \sqrt{1+\frac{1}{3}+\frac{1}{3}}}=\frac{3}{5}
$$

Therefore, the sine value of the dihedral angle is $\frac{4}{5}$.