13. The line $y=m x+b$ (where $|m|<1,|b|<1$) intersects the circle $x^{2}+y^{2}=1$ at points $P$ and $Q$, and intersects the hyperbola $x^{2}-y^{2}=1$ at points $R$ and $S$. If points $P$ and $Q$ trisect the segment $R S$, find the values of $m$ and $b$.
13. Let the coordinates of $P, Q, R, S$ be $\left(x_{i}, y_{i}\right), i=1,2,3,4$, from $\left\{\begin{array}{l}y=m x+b \\ x^{2}+y^{2}=1\end{array}\right.$, eliminating $y$, we get $\left(1+m^{2}\right) x^{2}+2 m b x+b^{2}-1=0$.
By Vieta's formulas, we have $x_{1}+x_{2}=\frac{-2 m b}{1+m^{2}}$,
From $\left\{\begin{array}{l}y=m x+b \\ x^{2}-y^{2}=1\end{array}\right.$, eliminating $y$, we get $\left(1-m^{2}\right) x^{2}-2 m b x-\left(b^{2}+1\right)=0$.
By Vieta's formulas, we have $x_{3}+x_{4}=\frac{2 m b}{1-m^{2}}$.
From (3), (4) we get $\frac{2 m b}{1+m^{2}}=\frac{2 m b}{1-m^{2}}$. Solving this, we get $m=0$ or $b=0$.
(1) When $m=0$, (1), (3) become $x^{2}+b^{2}-1=0, x^{2}-b^{2}-1=0$. Solving these, we get $x_{1,2}= \pm \sqrt{1-b^{2}}, x_{3,4}= \pm \sqrt{1+b^{2}}$, and then consider other aspects of the problem as required. From $|P Q|=\frac{1}{3}|R S|$, we get $\left|x_{1}-x_{2}\right|=\frac{1}{3}\left|x_{3}-x_{4}\right|$, thus $2 \sqrt{1-b^{2}}=\frac{2}{3} \sqrt{1+b^{2}}$. Solving this, we get $b= \pm \frac{2 \sqrt{5}}{5}$. After verification, $m=0, b= \pm \frac{2 \sqrt{5}}{5}$ satisfy the problem requirements.
(2) If $b=0$, similarly, we can find $m= \pm \frac{2 \sqrt{5}}{5}$.