Three. (20 points) As shown in Figure 2, let the pyramid be $P$ $A B C D$, with the base $A B C D$ being a rhombus, $P A \perp$ plane $A B C D, \angle A B C=60^{\circ}$, and $E, F$ being the midpoints of $B C, P C$ respectively.
(1) Prove: $A E \perp P D$;
(2) If $H$ is a moving point on $P D$, and the maximum angle formed by $E H$ with the plane $P A D$ has a tangent value of $\frac{\sqrt{6}}{2}$, find the cosine value of the dihedral angle $E-A F-C$.
(1) Since quadrilateral $ABCD$ is a rhombus, $\angle ABC = 60^{\circ}$, we know that $\triangle ABC$ is an equilateral triangle.
Since $E$ is the midpoint of $BC$, $AE \perp BC$.
Also, $BC \parallel AD$, so $AE \perp AD$.
Since $PA \perp$ plane $ABCD$ and $AE \subset$ plane $ABCD$, we have $PA \perp AE$. Since $PA \subset$ plane $PAD$ and $AD \subset$ plane $PAD$, and the intersection point of $PA$ and $AD$ is $A$, it follows that $AE \perp$ plane $PAD$.
Since $PD \subset$ plane $PAD$, therefore, $AE \perp PD$.
(2) Let $AB = 2$. Connect $AH$ and $EH$.
From (1), we know that $AE \perp$ plane $PAD$. Therefore, $\angle EHA$ is the angle between $EH$ and plane $PAD$.
In the right triangle $\triangle EAH$, since $AE = \sqrt{3}$, when $AH$ is the shortest, i.e., $AH \perp PD$, $\angle EHA$ 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 $AD = 2$, $\angle ADH = 45^{\circ}$.
Therefore, $PA = 2$.
The following are two methods to find the cosine value of the dihedral angle $E-AF-C$.
Solution 1 Since $PA \perp$ plane $ABCD$ and $PA \subset$ plane $PAC$, we have plane $PAC \perp$ plane $ABCD$. Draw $EO \perp AC$ at point $O$. Then $EO \perp$ plane $PAC$.
Draw $OS \perp AF$ at point $S$, and connect $ES$. Then $\angle ESO$ is the plane angle of the dihedral angle $E-AF-C$.
In the right triangle $\triangle AOE$,
$EO = AE \sin 30^{\circ} = \frac{\sqrt{3}}{2}$,
$AO = AE \cos 30^{\circ} = \frac{3}{2}$.
In the right triangle $\triangle ASO$,
$SO = AO \sin 45^{\circ} = \frac{3\sqrt{2}}{4}$.
Thus, $SE = \sqrt{EO^2 + SO^2} = \sqrt{\frac{3}{4} + \frac{9}{8}} = \frac{\sqrt{30}}{4}$.
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 dihedral angle is $\frac{\sqrt{15}}{5}$.
Solution 2 From (1), we know that $AE$, $AD$, and $AP$ are mutually perpendicular.
Establish a spatial rectangular coordinate system with $A$ as the origin as shown in Figure 3.

Since $E$ and $F$ are the midpoints of $BC$ and $PC$ respectively, we have
$A(0,0,0)$,
$B(\sqrt{3},-1,0)$,
$C(\sqrt{3}, 1,0)$,
$D(0,2,0)$, $P(0,0,2)$, $E(\sqrt{3}, 0,0)$,
$F\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 1\right)$.
Thus, $\overrightarrow{AE} = (\sqrt{3}, 0,0)$, $\overrightarrow{AF} = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}, 1\right)$.
Let the normal vector of plane $AEF$ be $m = (x_1, y_1, z_1)$.
Take $z_1 = 1$. Then $\boldsymbol{m} = (0,2,-1)$.
Since $BD \perp AC$, $BD \perp PA$, and $PA$ intersects $AC$ at point $A$, we have
$BD \perp$ plane $AFC$.
Thus, $\overrightarrow{BD}$ is the normal vector of plane $AFC$.
Since $\overrightarrow{BD} = (-\sqrt{3}, 3,0)$, we have
$\cos \langle\boldsymbol{m}, \overrightarrow{BD}\rangle = \frac{\boldsymbol{m} \cdot \overrightarrow{BD}}{|\boldsymbol{m}||\overrightarrow{BD}|}$
$= \frac{2 \times 3}{\sqrt{5} \times \sqrt{12}} = \frac{\sqrt{15}}{5}$.
Since the dihedral angle $E-AF-C$ is acute, the cosine value of the dihedral angle is $\frac{\sqrt{15}}{5}$.