14. (10 points) As shown in Figure 1, the quadrilateral pyramid $P-ABCD$, $PA \perp$ plane $ABCD$, and $PA=4$. The base $ABCD$ is a right trapezoid,
$$
\begin{array}{l}
\angle CDA = \angle BAD \\
= 90^{\circ}, AB = 2, CD \\
= 1, AD = \sqrt{2}, M, N
\end{array}
$$

are the midpoints of $PD$ and $PB$ respectively, and the intersection point of plane $MCN$ with $PA$ is $Q$.
(1) Find the length of $PQ$;
(2) Find the size of the dihedral angle formed by the section $MCN$ and the base $ABCD$;
(3) Find the distance from point $A$ to the plane $MCN$.
14. (1) Take the midpoint $E$ of $AP$, and connect $ED$. Then $ED \parallel CN$.
Take the midpoint of $EP$ as point $Q$, since $MQ \parallel ED$, hence $MQ \parallel CN$.

Therefore, points $M, N, C, Q$ are coplanar, and the intersection point $Q$ of plane $MCN$ with $AP$ is the quarter point of $AP$.
Thus, $PQ=1$.
(2) It is easy to prove that plane $MEN \parallel$ base $ABCD$. Therefore, the dihedral angle formed by the section $MCN$ and plane $MEN$ is the same as the dihedral angle formed by the section $MCN$ and base $ABCD$.
Since $PA \perp$ plane $ABCD$, it follows that
$PA \perp$ plane $MEN$.
Draw $EF \perp MN$ from $E$, with the foot of the perpendicular at $F$, and connect $QF$.
Then, by the theorem of three perpendiculars, we have $QF \perp MN$.
Thus, $\angle QFE$ is the plane angle of the dihedral angle formed by the section $MCN$ and plane $MEN$.
In the right triangle $MEN$,
$ME=\frac{\sqrt{2}}{2}, EN=1, MN=\frac{\sqrt{6}}{2}$.
Hence, $EF=\frac{\sqrt{3}}{3}$.
Therefore, $\tan \angle QFE=\sqrt{3}$.
Thus, $\angle QFE=\frac{\pi}{3}$.
(3) Since the midpoint of $EP$ is $Q$, and plane $MCN$ intersects $PA$ at point $Q$, the distance from point $A$ to plane $MCN$ is three times the distance from point $E$ to plane $MCN$.

From (2), we know that $MN \perp$ plane $QEF$. Therefore, plane $MCNQ$ is perpendicular to plane $QEF$ and their intersection is $QF$.

Draw $EH \perp QF$, with the foot of the perpendicular at $H$.
Then $EH \perp$ plane $MCNQ$, and $EH$ is the distance from point $E$ to plane $MCN$.
In the right triangle $EQF$,
$EF=\frac{\sqrt{3}}{3}, \angle QFE=\frac{\pi}{3}$.
Hence, $EH=\frac{1}{2}$.
Therefore, the distance from point $A$ to plane $MCN$ is $\frac{3}{2}$.