9. In rectangle $A B C D$, $A D=4, A B=3$, fold along diagonal $A C$ to form a right dihedral angle $B_{1}-A C-D$, find:
(1) the size of the dihedral angle $B_{1}-D C-A$;
(2) the distance between the skew lines $A B_{1}$ and $C D$.
9. (1) Draw $B_{1} E \perp A C$ at $E$. Since the dihedral angle $B_{1}-A C-D$ is a right dihedral angle, we know that $B_{1} E \perp$ plane $A C D$. Draw $E F \perp C D$ at $F$, and connect $B_{1} F$. By the theorem of three perpendiculars, we know that $B_{1} F \perp C D$, so $\angle B_{1} F E$ is the plane angle of the dihedral angle $B_{1}-D C-A$. By the area formula of a triangle, we know that $\frac{1}{2} B_{1} E \cdot A C=\frac{1}{2} A B_{1} \cdot B_{1} C$, thus $B_{1} E=\frac{12}{5}$, and then $C E=\sqrt{4^{2}-\left(\frac{12}{5}\right)^{2}}=\frac{16}{5}$. Noting that $E F / / A D$, hence $\triangle C E F \sim \triangle C A D$, thus $\frac{E F}{A D}=\frac{C E}{C A}=\frac{16}{25}$, getting $\quad E F=\frac{64}{25}$. Therefore, $\tan \angle B_{1} F E=\frac{B_{1} E}{E F}=\frac{15}{16}$, thus the size of the required dihedral angle is $\arctan \frac{15}{16}$.
(2) Noting that $A B / / C D$, hence the distance between the skew lines $A B_{1}$ and $C D$ equals the distance $h$ from point $D$ to the plane $A B_{1} B$. $V_{B_{1}-A B D}=\frac{1}{3} \times B_{1} E \times \frac{1}{2} \times A B \times A D=\frac{1}{3} \times \frac{12}{5} \times \frac{1}{2} \times 4 \times 3=\frac{24}{5}$. Since $B B_{1}^{2}=B E^{2}+B_{1} E^{2}=A B^{2}+A E^{2}-2 A B \cdot A E \cdot \cos \angle B A C+B_{1} E^{2}$
$$
=3^{2}+\left(5-\frac{16}{5}\right)^{2}-2 \times 3 \times\left(5-\frac{16}{5}\right) \times \frac{3}{5}+\left(\frac{12}{5}\right)^{2}=\frac{288}{25}
$$

In the isosceles $\triangle B A B_{1}$, the distance from point $A$ to $B B_{1}$ is $h_{1}=\sqrt{A B^{2}-\left(\frac{B B_{1}}{2}\right)^{2}}=\frac{3 \sqrt{17}}{5}$.
Thus, $V_{D-A B_{1} B}=\frac{1}{3} h \times \frac{1}{2} h_{1} \times B B_{1}=\frac{6 \sqrt{34}}{25} h$.
By $V_{D-A B_{1} B}=V_{B_{1}-A D B}$, we know that $h=\frac{10 \sqrt{34}}{17}$.
Therefore, the distance between the skew lines $A B_{1}$ and $C D$ is $\frac{10 \sqrt{34}}{17}$.