16 (12 points) In the tetrahedron $A-BCD$, $\triangle BCD$ and $\triangle ACD$ are both equilateral triangles with side length 2, and $\triangle BCD$ lies in plane $\alpha$. The side length $AB=\sqrt{3}$. Now, four of the eight labels numbered 1 to 8 are randomly attached to the four vertices, and the corresponding labels are denoted as $f(\eta)$ (where $\eta$ takes values $A, B, C, D$), and $E$ is a point on the side $AB$.
(1) Find the probability $P_{1}$ of the event “$f(C)+f(D)$ is even”;
(2) If $|BE|:|EA|=f(B):f(A)$, find the probability $P_{2}$ that the dihedral angle $E-CD-A$ is greater than $\frac{\pi}{4}$.
16 (1) Let $M_{1}$ denote the event that “both $f(C)$ and $f(D)$ are odd numbers”, and $M_{2}$ denote the event that “both $f(C)$ and $f(D)$ are even numbers”. According to the problem, we have
$$
\begin{array}{l}
P\left(M_{1}\right)=\frac{\mathrm{A}_{4}^{2}}{\mathrm{~A}_{8}^{2}}=\frac{4 \times 3}{8 \times 7}=\frac{3}{14}, \\
P\left(M_{2}\right)=\frac{\mathrm{A}_{4}^{2}}{\mathrm{~A}_{8}^{2}}=\frac{4 \times 3}{8 \times 7}=\frac{3}{14} .
\end{array}
$$

Let “$f(C)+f(D)$ is an even number” be the event $Q$, then $Q=M_{1}+M_{2}$, so
$$
P_{1}=P\left(M_{1}\right)+P\left(M_{2}\right)=\frac{3}{14} \times 2=\frac{3}{7} .
$$
(2) As shown in the figure, take the midpoint $F$ of $C D$, and connect $B F$, $A F$, and $E F$.

Since $\triangle B C D$ and $\triangle A C D$ are both equilateral triangles with side length 2, we have $A F \perp C D$ and $B F \perp C D$, thus $C D \perp$ plane $A B F$.

Therefore, $\angle A F E$ is the plane angle $\theta$ of the dihedral angle $E-C D-A$.
Also, $A F=B F=\sqrt{3}=A B$, so $\angle A F B=\frac{\pi}{3}$.
If $\theta=\frac{\pi}{4}$, then $\angle E F B=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$, at this time
$$
\begin{aligned}
\frac{|A E|}{|B E|} & =\frac{S_{\triangle A E F}}{S_{\triangle B E F}}=\frac{\frac{1}{2}|A F| \cdot|E F| \cdot \sin \frac{\pi}{4}}{\frac{1}{2}|B F| \cdot|E F| \cdot \sin \frac{\pi}{12}} \\
= & \frac{\sin \frac{\pi}{4}}{\sin \frac{\pi}{12}}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{6}-\sqrt{2}}{4}}=\sqrt{3}+1,
\end{aligned}
$$

Therefore, $\theta$ is greater than $\frac{\pi}{4}$, i.e., $\frac{f(A)}{f(B)}>\sqrt{3}+1$.
When $f(B)=1$, $f(A) \geqslant 3$, so $f(A)$ can take 6 values: $3,4,5, \cdots, 8$;
When $f(B)=2$, $f(A) \geqslant 6$, so $f(A)$ can take 3 values: $6, 7, 8$;
When $f(B) \geqslant 3$, $f(A) \geqslant 9$, so $f(A)$ does not exist.
Therefore, $P_{2}=\frac{9}{A_{8}^{2}}=\frac{9}{56}$.