14. Given the plane development diagram (as shown in Figure (2)) of the tetrahedron $P-ABC$ (as shown in Figure (1)), where quadrilateral $ABCD$ is a square with side length $2\sqrt{2}$, and $\triangle ABE$ and $\triangle BCF$ are both equilateral triangles, in the tetrahedron $P-ABC$:
(1) Prove: Plane $PAC \perp$ Plane $ABC$;
(2) If point $M$ is a point on edge $PA$ such that $\frac{PM}{MA}=\frac{1}{2}$, find the cosine value of the dihedral angle $P-BC-M$.
(1) Let the midpoints of $AB, AC$ be $G, H$, and connect $PG, PH, HG$. Then $PG \perp AB, HG \perp AB$
$\Rightarrow AB \perp$ plane $PHG \Rightarrow AB \perp PH$, similarly $BC \perp PH$.
Thus $PH \perp$ plane $ABC \Rightarrow$ plane $PAC \perp$ plane $ABC$.
(2) Establish a rectangular coordinate system with $\overrightarrow{HC}, \overrightarrow{HB}, \overrightarrow{HP}$ as the positive directions of the $x, y, z$ axes, respectively. Then $C(2,0,0), B(0,2,0), P(0,0,2)$, $A(-2,0,0), M\left(-\frac{2}{3}, 0, \frac{4}{3}\right)$, so $\overrightarrow{PC}=(2,0,-2), \overrightarrow{MC}=\left(\frac{8}{3}, 0,-\frac{4}{3}\right)$. Let the normal vectors of planes $PBC, MBC$ be $\boldsymbol{m}=\left(x_{1}, y_{1}, z_{1}\right), \boldsymbol{n}=\left(x_{2}, y_{2}, z_{2}\right)$, then $\left\{\begin{array}{l}x_{1}-y_{1}=0, \\ x_{1}-z_{1}=0\end{array} \Rightarrow \boldsymbol{m}=(1,1,1),\left\{\begin{array}{l}x_{1}-y_{1}=0, \\ 2 x_{1}-z_{1}=0\end{array} \Rightarrow \boldsymbol{n}=(1,1,2)\right.\right.$. Therefore, $\cos \langle\boldsymbol{m}, \boldsymbol{n}\rangle=\frac{4}{\sqrt{3} \cdot \sqrt{6}}=\frac{2 \sqrt{2}}{3}$. That is, the cosine value of the dihedral angle $P-BC-M$ is $\frac{2 \sqrt{2}}{3}$.