(13) As shown in the figure, the basic unit of a certain building can be approximately constructed as follows: First, draw a rhombus $ABCD$ on the ground plane $\alpha$, with side length $1$ and $\angle BAD=60^{\circ}$. Then, above $\alpha$, install two identical regular pyramids $P-ABD$ and $Q-CBD$ with $\triangle ABD$ and $\triangle CBD$ as their bases, respectively. It is known that $\angle APB=90^{\circ}$.
(1) Find the cosine value of the dihedral angle $P-BD-Q$;
(2) Find the distance from point $P$ to the plane $QBD$.
(13) (1) As shown in the figure, construct $P O_{1} \perp$ plane $A B C D$ at point $O_{1}$, and $Q O_{2} \perp$ plane $A B C D$ at point $O_{2}$. Since $P-A B D$ and $Q-C B D$ are both regular tetrahedrons, points $O_{1}$ and $O_{2}$ are the centers of $\triangle A B D$ and $\triangle C B D$, respectively. Also, since $O_{1} O_{2} \perp B D$, it follows that $B D \perp$ plane $P O_{1} O_{2} Q$, and thus $B D \perp P Q$. Taking the midpoint $R$ of $B D$, connect $P R$ and $Q R$, then $\angle P R Q$ is the plane angle of the dihedral angle $P-B D-Q$. It is easy to calculate that
$$
P R=Q R=\frac{1}{2}, P Q=O_{1} O_{2}=\frac{\sqrt{3}}{3} .
$$

Therefore,
$$
\begin{aligned}
\cos \angle P R Q & =\frac{P R^{2}+Q R^{2}-P Q^{2}}{2 P R \cdot Q R} \\
& =\frac{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}{2 \cdot \frac{1}{2} \cdot \frac{1}{2}} \\
& =\frac{1}{3} .
\end{aligned}
$$
(2) Construct $P H \perp R Q$ at point $H$. Since $B D \perp$ plane $P O_{1} O_{2} Q$ and $P H \subset$ plane $P O_{1} O_{2} Q$, it follows that $B D \perp P H$. Therefore, $P H \perp$ plane $B D Q$, and $P H$ is the required distance,
$$
P H=P R \sin \angle P R Q=\frac{1}{2} \sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{\sqrt{2}}{3} .
$$