20. (1) According to the problem, establish a spatial rectangular coordinate system $O-xyz$, then $O_{1}(0,1, \sqrt{3}), A(\sqrt{3}, 0,0), A_{1}(\sqrt{3}, 1, \sqrt{3}), B(0,2,0)$, hence $\overrightarrow{A B}=(-\sqrt{3}, 2,0), \overrightarrow{A A_{1}}=(0,1, \sqrt{3})$, clearly $\vec{k}=(0,0,1)$ is the normal vector of plane $OAB$, let $\vec{n}=(x_{1}, y_{1}, z_{1})$ be the normal vector of plane $O_{1}AB$.
From $\left\{\begin{array}{l}\vec{n} \cdot \overrightarrow{A B}=0 \\ \vec{n} \cdot \overrightarrow{A A_{1}}=0\end{array} \Rightarrow\left\{\begin{array}{l}-\sqrt{3} x_{1}+2 y_{1}=0, \\ y_{1}+\sqrt{3} z_{1}=0 .\end{array}\right.\right.$
Let $x_{1}=2$, we get $\vec{n}=(2, \sqrt{3},-1)$,
then $\cos \langle\vec{n}, \vec{k}\rangle=|\vec{n} \cdot \vec{k}| /|\vec{n}||\vec{k}|=\frac{\sqrt{2}}{4}$.
Therefore, the size of the dihedral angle $O-AB-O_{1}$ is $\arccos \frac{\sqrt{2}}{4}$.
(2) From (1), we get $\overrightarrow{B A_{1}}=(\sqrt{3},-1, \sqrt{3})$.
Let $\vec{m}=(x_{2}, y_{2}, z_{2})$, and $\vec{m} \perp \overrightarrow{OA}, \vec{m} \perp A_{1}B$,

From $\left\{\begin{array}{l}\vec{m} \cdot \overrightarrow{O A}=0 \\ \vec{m} \cdot \overrightarrow{B A_{1}}=0\end{array} \Rightarrow\left\{\begin{array}{l}\sqrt{3} x_{2}=0, \\ \sqrt{3} x_{2}-y_{2}+\sqrt{3} z_{2}=0 .\end{array}\right.\right.$
Let $z_{2}=1$, we get $\vec{m}=(0, \sqrt{3}, 1)$.
Thus, $d=\left|\overrightarrow{A A_{1}} \cdot \vec{m}\right| /|\vec{m}|=\sqrt{3}$.
20. (2002 Shanghai Spring College Entrance Examination) Triangular prism $O A B-O A_{1} B_{1}$, plane $O B B_{1} O_{1} \perp O A B, \angle O_{1} O B=$ $60^{\circ}, \angle A O B=90^{\circ}$, and $O B=O O_{1}=2, O A=\sqrt{3}$. Find:
(1) The size of the dihedral angle $O_{1}-A B-O$;
(2) The distance $d$ between the skew lines $A_{1} B$ and $O A$ (this question is adapted from the original).