(14) As shown in the figure, the edge lengths of the oblique triangular prism $A B C-A_{1} B_{1} C_{1}$ are all $a$, the side face $B_{1} C_{1} C B \perp$ the base $A B C$, and $A C_{1} \perp B C$.
(1). Find the distance between the skew lines $A A_{1}$ and $B_{1} C_{1}$;
(2) Find the degree of the dihedral angle formed by the side face $A_{1} B_{1} B A$ and the base $A B C$.
14 (1) As shown in the figure, take the midpoint $D$ of $B C$, and connect $A D, C_{1} D$. Since $A D \perp B C, A C_{1} \perp B C$, it follows that $B C \perp$ plane $A D C_{1}$, hence $C_{1} D \perp B C$. Because plane $B_{1} C_{1} C B \perp$ base plane $A B C$, we have $C_{1} D \perp$ plane $A B C$.

From $A D \perp B C$, we know $A D \perp$ plane $B_{1} C_{1} C B$. Since $A A_{1} / / C C_{1}$, it follows that $A A_{1} / /$ plane $B_{1} C_{1} C B$. Therefore, the distance between the skew lines $A A_{1}$ and $B_{1} C_{1}$ is $A D=\frac{\sqrt{3}}{2} a$.
(2) As shown in the figure, draw $B_{1} O \perp B C$, intersecting $B C$ at point $O$, then $B_{1} O \perp$ base plane $A B C$. Draw $O E \perp A B$, intersecting $A B$ at point $E$, and connect $B_{1} E$, then $\angle B_{1} E O$ is supplementary to the dihedral angle we are looking for. It is easy to see that $B_{1} O=C_{1} D=\frac{\sqrt{3}}{2} a, O B=\frac{a}{2}$, $O E=\frac{\sqrt{3}}{4} a$. Therefore,
$$
\tan \angle B_{1} E O=\frac{B_{1} O}{O E}=\frac{\frac{\sqrt{3}}{2} a}{\frac{\sqrt{3}}{4} a}=2,
$$

so $\angle B_{1} E O=\arctan 2$. Therefore, the dihedral angle between side face $A_{1} B_{1} B A$ and base plane $A B C$ is $\pi-\arctan 2$.