3. (20 points) As shown in Figure 1, in rectangle $ABCD$, $AB = \sqrt{3}$, $BC = a$. Also, $PA \perp$ plane $ABCD$, $PA = 4$.
(1) If there exists a point $Q$ on side $BC$ such that $PQ \perp QD$, find the range of values for $a$;
(2) When there is exactly one point $Q$ on $BC$ such that $PQ \perp QD$, find the angle between the skew lines $AQ$ and $PD$;
(3) If $a = 4$, and $PQ \perp QD$, find the size of the dihedral angle $A - PD - O$.
(1) Let $B Q=t$, then
$P Q^{2}=19+t^{2}, \varphi()^{2}=3+(a-t)^{2}$,
$P I)^{2}=16+a^{2}$.
If $P Q \perp Q D$, we get $19+t^{2}+3+(a-1)^{2}=16+a^{2}$,
thus $t^{2}-a t+3=0$.
From $\Delta=a^{2}-12 \geqslant 0$, solving gives $a \geqslant 2 \sqrt{3}$.
(2) For the minimum value of $B C$: there exists a unique point $Q$, such that $P Q \perp Q D$.
If $\Delta=a^{2}-12=0$, solving gives $a=2 \sqrt{3}, t=\sqrt{3}$.
As shown in the figure, take the midpoint $R$ of $A D$, the midpoint $S$ of $P A$, and connect $R S$, $R C$. Then $R S / / P D, R C / / A Q$. Thus, $\angle S R C$ is the angle formed by the skew lines $A Q$ and $P D$.
Also, $R S=\frac{1}{2} P l=\sqrt{7}, R C=A Q=\sqrt{6}$,
$S C=\sqrt{S A^{2}+A C^{2}}=\sqrt{19}$.
Therefore, $\cos \angle S R C=\frac{R S^{2}+R C^{2}-S C^{2}}{2 R S \cdot R C}=-\frac{\sqrt{42}}{14}$.
Hence, the angle formed by the skew lines $A Q$ and $P D$ is $\arccos \frac{\sqrt{42}}{14}$.
(3) Draw $Q M / / C D$ intersecting $A D$ at point $M$, then $Q M \perp A l, A M=t$.
Since $P A \perp$ plane $A B C D$, then $P A \perp Q M$.
Also, $Q M \perp A D$, then $Q M \perp$ plane $P A D$.
Draw $M N \perp P D$ at point $N$, and connect $N Q$.
By the theorem of three perpendiculars, $Q N \perp P D$.
Thus, $\angle M N Q$ is the plane angle of the dihedral angle $A-P D-Q$.
In the right triangle $\triangle P A D$, by $\frac{M N}{P A}=\frac{M D}{P D}$, we have $M N=\frac{4-1}{\sqrt{2}}$.
In the right triangle $\triangle P Q D$, by $\frac{N Q}{P Q}=\frac{D Q}{P D}$, we have
$$
N Q=\frac{\sqrt{19+t^{2}} \times \sqrt{3+(4-t)^{2}}}{4 \sqrt{2}} .
$$

Then $\cos \angle M N Q=\frac{M N}{N Q}=\frac{4(4-1)}{\sqrt{19+t^{2}} \times \sqrt{3+(4-1)^{2}}}$.
From equation (1), we get $t=1$ or $t=3$.
When $t=1$, $\cos \angle M N Q=\frac{\sqrt{15}}{5}$;
When $t=3$, $\cos \angle M N Q=\frac{\sqrt{7}}{7}$.
Therefore, the size of the dihedral angle $A-P D-Q$ is
$\arccos \frac{\sqrt{15}}{5}$ or $\arccos \frac{\sqrt{7}}{7}$.