13. As shown in Figure 1, the basic unit of a certain building can be approximately constructed as follows: First, construct 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 $O-CBD$ with $\triangle ABD$ and $\triangle CBD$ as their bases, respectively, and $\angle APB=90^{\circ}$.
(1) Find the cosine value of the dihedral angle $P-BD-Q$; (2) Find the distance from point $P$ to the plane $QBD$.
Three, 13. (1) As shown in Figure 5, draw $P O_{1} \perp$ 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, $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ are the centers of the equilateral triangles $\triangle A B D$ and $\triangle C B D$, respectively.

Also, $O_{1} O_{2} \perp B D$, so $B D \perp$ plane $P O_{1} O_{2} Q$.
Therefore, $B D \perp P Q$.
(2) As shown in Figure 5, take the midpoint $R$ of $B D$, and connect $P R$ and $Q R$. $\angle P R Q$ is the plane angle of the dihedral angle $P-B D-Q$. It is easy to calculate that
$$
\begin{array}{l}
P R=Q R=\frac{1}{2}, P Q=O_{1} O_{2}=\frac{\sqrt{3}}{3} . \\
\text { Hence } \cos \angle P R Q=\frac{P R^{2}+Q R^{2}-P Q^{2}}{2 P R \cdot Q R} \\
=\frac{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}{2 \times \frac{1}{2} \times \frac{1}{2}}=\frac{1}{3} .
\end{array}
$$
(3) As shown in Figure 6, draw $P H \perp R Q$ at point $H$.

Since $B D \perp$ plane $P O_{1} O_{2} Q$ and $\mathrm{PH} \subset$ plane $\mathrm{P} O_{1} O_{2} Q$, it follows that
$B D \perp P H$.
Therefore, $P H \perp$ plane $B D Q$.
Thus, $P H$ is the required length, and
$$
\begin{array}{r}
P H=P R \sin \angle P R Q \\
=\frac{1}{2} \sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{\sqrt{2}}{3} .
\end{array}
$$