In a regular quadrilateral pyramid S-ABCD, the height is $a$, and the side length of the base is $2a$.
(1) Find the angle between the base and a side face;
(2) Find the distance from B to the side edge SC;
(3) Find the angle between two adjacent side faces.
(1) Draw $O F \perp B C$ at $F$,

Connect $\mathrm{SF}$, then $\mathrm{SF} \perp \mathrm{BC}$ (by the theorem of three perpendiculars).
Let $\angle \mathrm{SFO}=\alpha$, then $\alpha$ is the angle we are looking for.
$$
\begin{array}{l}
\because \text { OF }=\mathrm{SO}=\mathrm{a} \\
\therefore \alpha=45^{\circ}, \text { i.e., }
\end{array}
$$

The angle between the base and the side face is $45^{\circ}$.
(2) Draw $B E \perp S C$, then $\mathrm{BE}$ is the distance we are looking for.
$$
\begin{array}{l}
\text { In Rt } \triangle \mathrm{SOF}, \mathrm{SF}=\sqrt{2} \mathrm{a}, \\
\text { In Rt } \triangle \mathrm{SFC}, \mathrm{SC}=\sqrt{\mathrm{SF}^{2}+\mathrm{FC}^{2}} \\
\quad=\sqrt{\left(\sqrt{2 \mathrm{a})^{2}} + \mathrm{a}^{2}\right.}=\sqrt{3} \mathrm{a}, \\
\text { Also, } \because B F \cdot \mathrm{SC}=\mathrm{SF} \cdot \mathrm{BC}, \\
\mathrm{BE}=\frac{\mathrm{SF} \cdot \mathrm{BC}}{\mathrm{SC}}=\frac{\sqrt{2} \cdot 2 \mathrm{a}}{\sqrt{3 \mathrm{a}}}=\frac{2 \sqrt{6}}{3} \mathrm{a} .
\end{array}
$$
(3) Connect $\mathrm{DE}$, by $\triangle \mathrm{BCF} \cong \triangle D C E$ we get $D E \perp C$. Let $\angle B E D=\beta$, then $\beta$ is the angle between two adjacent side faces. In $\triangle B E D$, by the cosine rule, we get
$$
\begin{array}{l}
\cos \beta=\frac{\mathrm{BE}^{2}+\mathrm{DE}^{2}-\mathrm{DB}^{2}}{2 \mathrm{BE} \cdot \mathrm{DE}}=\frac{2 \mathrm{BE}^{2}-\mathrm{DB}^{2}}{2 \mathrm{BE}^{2}} . \\
=\frac{2\left(\frac{2 \sqrt{6}}{3} \mathrm{a}^{2}-(2 \sqrt{2} \mathrm{a})^{2}\right.}{2\left(\frac{2 \sqrt{6}}{3} \mathrm{a}\right)^{2}}=-\frac{1}{2}, \\
\therefore \beta=120^{\circ} .
\end{array}
$$

The angle between two adjacent side faces is 120.