16. (12 points) In the tetrahedron $A-BCD$, it is known that $\triangle BCD$ and $\triangle ACD$ are both equilateral triangles with side lengths of 2, and $\triangle BCD$ lies in plane $\alpha$. The side length $AB=\sqrt{3}$. Now, four of the eight labels numbered $1 \sim 8$ 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 $|BE|=\frac{f(B)}{f(A)}$, find the probability $P_{2}$ of the event “the dihedral angle $E-CD-A$ is greater than $\frac{\pi}{4}$”.
16. (1) Let $M_{1}$ denote the event that “$f(C)$ and $f(D)$ are both odd numbers,” and $M_{2}$ denote the event that “$f(C)$ and $f(D)$ are both even numbers.”
From 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 $Q=M_{1}+M_{2}$.
Thus, $P_{1}=P\left(M_{1}\right)+P\left(M_{2}\right)=\frac{3}{14} \times 2=\frac{3}{7}$.
(2) As shown in Figure 7, take the midpoint $F$ of side $C D$, 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
$$
\begin{array}{l}
A F \perp C D, \\
B F \perp C D .
\end{array}
$$

Therefore,
$C D \perp$ plane $A B F$.
Thus, $\angle A F E$ is the plane angle $\theta$ of the dihedral angle $E-C D-A$.
Also, $A F=B F=\sqrt{3}=A B$, so $\angle A F B=\frac{\pi}{3}$.
Hence, $\frac{f(A)}{f(B)}=\frac{|A E|}{|B E|}=\frac{\sin \angle A F E}{\sin \angle B F E}$
$$
=\frac{\sin \theta}{\sin \left(\frac{\pi}{3}-\theta\right)}>\frac{\sin \frac{\pi}{4}}{\sin \frac{\pi}{12}}=\sqrt{3}+1 \text {. }
$$

When $f(B)=1$, $f(A) \geqslant 3$, so $f(A)$ can take the values $3,4, \cdots, 8$, a total of six values;

When $f(B)=2$, $f(A) \geqslant 6$, so $f(A)$ can take the values $6,7,8$, a total of three values;
When $f(B) \geqslant 3$, $f(A) \geqslant 9$, so $f(A)$ does not exist. In summary, $P_{2}=\frac{9}{A_{8}^{2}}=\frac{9}{56}$.