79. Place two congruent triangular boards as shown in Figure (1), where $\angle A_{1} C B_{1}=\angle A C B=90^{\circ}, \angle A_{1}=\angle A=30^{\circ}$. Rotate $\triangle A_{1} B_{1} C$ in Figure (1) $45^{\circ}$ clockwise around point $C$ to get Figure (2). Point $P_{1}$ is the intersection of $A_{1} C$ and $A B$, and point $Q$ is the intersection of $A_{1} B_{1}$ and $B C$. Given that $A P_{1}=2$, find:
(1) The length of $C Q$;
(2) If $S_{C B_{1} Q}: S_{C P_{1} A}=1: k$, find the value of $k$.
Solution: (1) According to the given conditions, we can prove that $\mathrm{D} B C P_{1} @ \mathrm{D} B_{1} C Q$, thus
$$
C P_{1}=C Q \text { . }
$$

Construct $P_{1} H^{\wedge} A C$ at point $H$, as shown in the figure.
In $R t \mathrm{D} A H P_{1}$, $Đ A=30^{\circ}, A P_{1}=2$, so $P_{1} H=1$.
In $R t \mathrm{D} P_{1} H C$, $P_{1} C H=45^{\circ}$, so $C P_{1}=\sqrt{2} P_{1} H=\sqrt{2}$.
Since
$$
C Q=C P_{1},
$$

we have
$$
C Q=\sqrt{2} \text { . }
$$
(2) From (1), we know $A C=1+\sqrt{3}$, in $R t \mathrm{D} A B C$, $Đ A=30^{\circ}$,

so
$$
B C=\frac{\sqrt{3}}{3} A C=\frac{\sqrt{3}}{3}(1+\sqrt{3})=\frac{\sqrt{3}}{3}+1, \quad A B=2 B C=\frac{2 \sqrt{3}}{3}+2 \text {, }
$$

thus
$$
P_{1} B=A B-P_{1} A=\left(\frac{2 \sqrt{3}}{3}+2\right)-2=\frac{2 \sqrt{3}}{3},
$$

and since
$$
A P_{1}=2,
$$

we have
$$
S_{C B P_{1}}: S_{C A P_{1}}=P_{1} B: P_{1} A=\frac{2 \sqrt{3}}{3}: 2=\sqrt{3}: 3 \text { . }
$$

Therefore,
$$
k=\sqrt{3} \text { . }
$$