14. As shown in the figure, in the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a rectangle, the side edge $PA \perp$ the base $ABCD$, $AB=\sqrt{3}$, $BC=1$, $PA=2$, and $E$ is the midpoint of $PD$.
(1) Find the cosine value of the angle formed by the lines $AC$ and $PB$;
(2) Find a point $N$ in the side face $PAB$ such that $NE \perp$ the plane $PAC$, and determine the distances from point $N$ to $AB$ and $AP$.
14. Solution 1 (1) As shown in the figure, establish a spatial rectangular coordinate system with $A$ as the origin,

then the coordinates of $A, B, C, D, P, E$ are
$$
B(\sqrt{3}, 0,0), C(\sqrt{3}, 1,0), D(0,1,0), P(0,0,2), E\left(0, \frac{1}{2}, 1\right) \text {, }
$$

Thus, $\overrightarrow{A C}=(\sqrt{3}, 1,0), \overrightarrow{P B}=(\sqrt{3}, 0,-2)$.
Let the angle between $\overrightarrow{A C}$ and $\overrightarrow{P B}$ be $\theta$, then
$$
\cos \theta=\frac{\overrightarrow{A C} \cdot \overrightarrow{P B}}{|\overrightarrow{A C}| \cdot|\overrightarrow{P B}|}=\frac{3}{2 \sqrt{7}}=\frac{3 \sqrt{7}}{14},
$$

Therefore, the cosine value of the angle formed by $A C$ and $P B$ is $\frac{3 \sqrt{7}}{14}$.
(2) Since point $N$ is on the side face $P A B$, we can set the coordinates of $N$ as $(x, 0, z)$, then
$\overrightarrow{N E}=\left(-x, \frac{1}{2}, 1-z\right)$, and since $N E \perp$ plane $P A C$, we have $\left\{\begin{array}{l}\overrightarrow{N E} \cdot \overrightarrow{A P}=0, \\ \overrightarrow{N E} \cdot \overrightarrow{A C}=0 .\end{array}\right.$,
which is $\left\{\begin{array}{l}\left(-x \cdot \frac{1}{2}, 1-z\right) \cdot(0,0,2)=0, \\ \left(-x, \frac{1}{2}, 1-z\right) \cdot(\sqrt{3}, 1,0)=0 .\end{array}\right.$,
simplifying to $\left\{\begin{array}{l}z-1=0, \\ -\sqrt{3} x+\frac{1}{2}=0 .\end{array}\right.$ Therefore, $\left\{\begin{array}{l}x=\frac{\sqrt{3}}{6} \\ z=1\end{array}\right.$
Thus, the coordinates of $N$ are $\left(\frac{\sqrt{3}}{6}, 0,1\right)$, and the distances from $N$ to $A B$ and $A P$ are $1, \frac{\sqrt{3}}{6}$, respectively.
Solution 2 (1) Let $A C \cap B D=O$, connect $O E$, then $O E / / P B$,
so $\angle E O A$ is the angle formed by $A C$ and $P B$ or its supplementary angle.
In $\triangle A O E$, $A O=1, O E=\frac{1}{2} P B=\frac{\sqrt{7}}{2}, A E=\frac{1}{2} P D=\frac{\sqrt{5}}{2}$,
so $\cos \angle E O A=\frac{1+\frac{7}{4}-\frac{5}{4}}{2 \times \frac{\sqrt{7}}{2} \times 1}=\frac{3 \sqrt{7}}{14}$.
Thus, the cosine value of the angle formed by $A C$ and $P B$ is $\frac{3 \sqrt{7}}{14}$.
(2) In the plane $A B C D$, draw a perpendicular from $D$ to $A C$ intersecting $A B$ at $F$, then $\angle A D F=\frac{\pi}{6}$.

In the right triangle $\triangle D A F$, $D F=\frac{A D}{\cos \angle A D F}=\frac{2 \sqrt{3}}{3}, A F=A D \cdot \tan \angle A D F=\frac{\sqrt{3}}{3}$.
Connect $P F$, let $N$ be the midpoint of $P F$, connect $N E$, then $N E / / D F$,
since $D F \perp A C, D F \perp P A$, so $D F \perp$ plane $P A C$, thus $N E \perp$ plane $P A C$.
Therefore, the distance from $N$ to $A B$ is $=\frac{1}{2} A P=1$, and the distance from $N$ to $A P$ is $=\frac{1}{2} A F=\frac{\sqrt{3}}{6}$.