(10 (20 points) As shown in the figure, given a pyramid $P-ABCD$, the base $ABCD$ is a rhombus, $PA \perp$ plane $ABCD$, $\angle ABC=60^{\circ}$, and $E$, $F$ are the midpoints of $BC$, $PC$ respectively.
(1) Prove: $AE \perp PD$;
(2) If $H$ is a moving point on $PD$, the maximum angle formed by $EH$ with the plane $PAD$ has a tangent value of $\frac{\sqrt{6}}{2}$, find the cosine value of the dihedral angle $E-AF-C$.
(10) (1) Since quadrilateral $ABCD$ is a rhombus, $\angle ABC=60^{\circ}$, it follows that $\triangle ABC$ is an equilateral triangle.

Since $E$ is the midpoint of $BC$, $AE \perp BC$. Also, since $BC \parallel AD$, it follows that $AE \perp AD$.
Since $PA \perp$ plane $ABCD$, and $AE \subset$ plane $ABCD$, it follows that $PA \perp AE$.
Since $PA \subset$ plane $PAD$, $AD \subset$ plane $PAD$, and $PA \cap AD=A$, it follows that $AE \perp$ plane $PAD$. Since $PD \subset$ plane $PAD$, it follows that $AE \perp PD$.
(2) Let $AB=2$, and $H$ be any point on $PD$, connect $AH$, $EH$.
From (1), we know $AE \perp$ plane $PAD$, so $\angle EHA$ is the angle between $EH$ and plane $PAD$.

In the right triangle $\triangle EAH$, $AE=\sqrt{3}$, so when $AH$ is the shortest, $\angle EHA$ is the largest, i.e., when $AH \perp PD$, $\angle EHA$ is the largest. At this time,
$$
\tan \angle EHA=\frac{AE}{AH}=\frac{\sqrt{3}}{AH}=\frac{\sqrt{6}}{2},
$$

therefore $AH=\sqrt{2}$. Since $AD=2$, it follows that $\angle ADH=45^{\circ}$, so $PA=2$.
Solution 1 Since $PA \perp$ plane $ABCD$, and $PA \subset$ plane $PAC$, it follows that plane $PAC \perp$ plane $ABCD$. Draw $EO \perp AC$ at $O$, then $EO \perp$ plane $PAC$, draw $OS \perp AF$ at $S$, connect $ES$, then $\angle ESO$ is the plane angle of the dihedral angle $E-AF-C$.
In the right triangle $\triangle AOE$,
$$
\begin{array}{l}
EO=AE \cdot \sin 30^{\circ}=\frac{\sqrt{3}}{2}, \\
AO=AE \cdot \cos 30^{\circ}=\frac{3}{2},
\end{array}
$$

Since $F$ is the midpoint of $PC$, in the right triangle $\triangle ASO$, $SO=AO \cdot \sin 45^{\circ}=\frac{3 \sqrt{2}}{4}$, and
$$
SE=\sqrt{EO^{2}+SO^{2}}=\sqrt{\frac{3}{4}+\frac{8}{9}}=\frac{\sqrt{30}}{4} .
$$

In the right triangle $\triangle ESO$,
$$
\cos \angle ESO=\frac{SO}{SE}=\frac{\frac{3 \sqrt{2}}{4}}{\frac{\sqrt{30}}{4}}=\frac{\sqrt{15}}{5},
$$

thus the cosine value of the required dihedral angle is $\frac{\sqrt{15}}{5}$.
Solution 2 From (1), we know $AE, AD, AP$ are mutually perpendicular, with $A$ as the origin, establish the spatial rectangular coordinate system as shown in the figure, and since $E, F$ are the midpoints of $BC, PC$ respectively, so $A(0,0,0), B(\sqrt{3},-1,0)$, $C(\sqrt{3}, 1,0), D(0,2,0), P(0,0,2)$, $E(\sqrt{3}, 0,0), F\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 1\right)$, so $\overrightarrow{AE}=$ $(\sqrt{3}, 0,0), \overrightarrow{AF}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 1\right)$.

Let the normal vector of plane $AEF$ be $\vec{m}=\left(x_{1}, y_{1}, z_{1}\right)$, then $\left\{\begin{array}{l}\vec{m} \cdot \overrightarrow{AE}=0, \\ \vec{m} \cdot \overrightarrow{AF}=0,\end{array}\right.$ therefore
$$
\left\{\begin{array}{l}
\sqrt{3} x_{1}=0, \\
\frac{\sqrt{3}}{2} x_{1}+\frac{1}{2} y_{1}+z_{1}=0 .
\end{array}\right.
$$

Take $z_{1}=1$, then $\vec{m}=(0,2,-1)$.
Since $BD \perp AC, BD \perp PA, PA \cap AC=A$, it follows that $BD \perp$ plane $AFC$, so $\overrightarrow{BD}$ is a normal vector of plane $AFC$.
Since $\overrightarrow{BD}=(-\sqrt{3}, 3,0)$, it follows that
$$
\cos \langle\vec{m}, \overrightarrow{BD}\rangle=\frac{\vec{m} \cdot \overrightarrow{BD}}{|\vec{m}| \cdot|\overrightarrow{BD}|}=\frac{2 \times 3}{\sqrt{5} \times \sqrt{12}}=\frac{\sqrt{15}}{5} \text {. }
$$

Since the dihedral angle $E-AF-C$ is acute, the cosine value of the required dihedral angle is $\frac{\sqrt{15}}{5}$.