2. Given a regular triangular prism $A B C-A^{\prime} B^{\prime} C^{\prime}$, points $D$ and $E$ are on the side edges $B B^{\prime}$ and $C C^{\prime}$ respectively, and $E C=B C=2 B D$. A section is made through $A$, $D$, and $E$. If $A A^{\prime}=2 A B$,
(1) find the angle between the section and the base;
(2) find the volume ratio of the two parts the section divides the prism into.
2. (1) Extend $E D$ to intersect the extension of $C B$ at $F$,

From $D B / / E C, B D=\frac{1}{2} E C$, we get $F B=B C=A B$. Thus, $\angle B A F=\angle B F A=30^{\circ}$, so $\angle F A C=90^{\circ}$.

Therefore, $A F \perp A E$, which means $\angle E A C$ is the plane angle of the dihedral angle we are looking for. In the right triangle $\triangle A E C$, $E C=A C$, hence $\angle E A C=45^{\circ}$.
Another solution: We can use $\cos \theta=S_{\triangle A B C} / S_{\triangle A D E}$ to find it.
(2) Let $A B=a$, then $A A^{\prime}=2 a, B D=\frac{a}{2}, E C=a$, thus
$$
\frac{V_{A-B C D E}}{V_{A B C-A^{\prime} B^{\prime} C}}=\frac{2}{3} \cdot \frac{S_{B C D E}}{S_{B C C B^{\prime}}}=\frac{1}{4} \text {, so } \frac{V_{A D E-A^{\prime} B^{\prime} C^{\prime}}}{V_{A-B D E}}=3 \text {. }
$$