14. (13 points) In the plane development diagram (Figure 2) of the tetrahedron $P-ABC$ (as shown in Figure 1), quadrilateral $ABCD$ is a square with side length $2\sqrt{2}$, and $\triangle ABE$, $\triangle BCF$ are both equilateral triangles. In the tetrahedron $P-ABC$,
(1) Prove: plane $PAC \perp$ plane $ABC$;
(2) If $M$ is a point on edge $PA$ such that $\frac{PM}{MA}=\frac{1}{2}$, find the cosine value of the dihedral angle $P-BC-M$.
14. (1) Let the midpoint of $AC$ be $O$, and connect $BO, PO$.

From the problem, we have
$$
\begin{array}{l}
PA=PB=PC=2 \sqrt{2}, PO=2, \\
AO=BO=CO=2 .
\end{array}
$$

In $\triangle PAC$, since $PA=PC$ and $O$ is the midpoint of $AC$, we get $PO \perp AC$.

In $\triangle POB$, since $PO=2, OB=2, PB=2 \sqrt{2}$, and $PO^2 + OB^2 = PB^2$, then $PO \perp OB$.

Since $AC$ and $OB$ intersect at point $O$, and $AC, OB \subset$ plane $ABC$, therefore, $PO \perp$ plane $ABC$.
Also, $PO \subset$ plane $PAC$, hence
plane $PAC \perp$ plane $ABC$.
(2) Since $PO \perp$ plane $ABC$ and $OB \perp AC$, we get $PO \perp OB, PO \perp OC$.
Thus, we can establish a spatial rectangular coordinate system with the lines through $OC, OB, OP$ as the $x$-axis, $y$-axis, and $z$-axis, respectively, as shown in Figure 4.
Then $O(0,0,0), C(2,0,0), B(0,2,0)$,
$$
\begin{array}{l}
A(-2,0,0), P(0,0,2), M\left(-\frac{2}{3}, 0, \frac{4}{3}\right), \\
\overrightarrow{BC}=(2,-2,0), \overrightarrow{PC}=(2,0,-2), \\
\overrightarrow{MC}=\left(\frac{8}{3}, 0,-\frac{4}{3}\right) .
\end{array}
$$

Let the normal vector of plane $MBC$ be $\boldsymbol{m}=\left(x_{1}, y_{1}, z_{1}\right)$.
Then from $\left\{\begin{array}{l}\boldsymbol{m} \cdot \overrightarrow{BC}=0, \\ \boldsymbol{m} \cdot \overrightarrow{MC}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}-y_{1}=0, \\ 2 x_{1}-z_{1}=0 \text {. }\end{array}\right.\right.$
Let $x_{1}=1$, we get $y_{1}=1, z_{1}=2$, i.e., $\boldsymbol{m}=(1,1,2)$.
Let the normal vector of plane $PBC$ be $\boldsymbol{n}=\left(x_{2}, y_{2}, z_{2}\right)$.
Then from $\left\{\begin{array}{l}\boldsymbol{n} \cdot \overrightarrow{BC}=0, \\ \boldsymbol{n} \cdot \overrightarrow{PC}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}-y_{2}=0 \text {, } \\ x_{2}-z_{2}=0 .\end{array}\right.\right.$
Let $x_{2}=1$, we get $y_{2}=1, z_{2}=1$, i.e., $\boldsymbol{n}=(1,1,1)$.
Thus, $\cos \langle\boldsymbol{n}, \boldsymbol{m}\rangle=\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}||\boldsymbol{n}|}=\frac{4}{\sqrt{18}}=\frac{2 \sqrt{2}}{3}$.
From Figure 4, the cosine value of the dihedral angle $P-BC-M$ is
$$
\text { } \frac{2 \sqrt{2}}{3} \text {. }
$$