8.399. $\left\{\begin{array}{l}x+y=\frac{\pi}{4} \\ \tan x \tan y=\frac{1}{6}\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \cos y \neq 0 .\end{array}\right.$

From the condition we have

$$
\begin{aligned}
& x=\frac{\pi}{4}-y, \operatorname{tg}\left(\frac{\pi}{4}-y\right) \operatorname{tg} y=\frac{1}{6} \Rightarrow \frac{\left(\operatorname{tg} \frac{\pi}{4}-\operatorname{tg} y\right) \operatorname{tg} y}{1+\operatorname{tg} \frac{\pi}{4} \operatorname{tg} y}=\frac{1}{6} \\
& \frac{(1-\operatorname{tg} y) \operatorname{tg} y}{1+\operatorname{tg} y}=\frac{1}{6}, 6 \operatorname{tg}^{2} y-5 \operatorname{tg} y+1=0
\end{aligned}
$$

Solving the last equation as a quadratic equation in $\operatorname{tg} y$, we get $\operatorname{tg} y_{1}=\frac{1}{2}, \operatorname{tg} y_{2}=\frac{1}{3}$. Substituting these values into the second equation of the system, we find $\operatorname{tg} x_{1}=\frac{1}{3}, \operatorname{tg} x_{2}=\frac{1}{2}$. Hence,

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x _ { 1 } = \operatorname { a r c t g } \frac { 1 } { 3 } + \pi k , } \\
{ y _ { 1 } = \frac { \pi } { 4 } - \operatorname { a r c t g } \frac { 1 } { 3 } - \pi k ; }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x_{2}=\operatorname{arctg} \frac{1}{2}+\pi n, \\
y_{2}=\frac{\pi}{4}-\operatorname{arctg} \frac{1}{2}-\pi n ;
\end{array} k, n \in Z .\right.\right. \\
& \text { Answer: }\left\{\begin{array} { l } 
{ x _ { 1 } = \operatorname { a r c t g } \frac { 1 } { 3 } + \pi k , } \\
{ y _ { 1 } = \frac { \pi } { 4 } - \operatorname { a r c t g } \frac { 1 } { 3 } - \pi k ; }
\end{array} \left\{\begin{array}{l}
x_{2}=\operatorname{arctg} \frac{1}{2}+\pi n, \\
y_{2}=\frac{\pi}{4}-\operatorname{arctg} \frac{1}{2}-\pi n ;
\end{array} k, n \in Z .\right.\right.
\end{aligned}
$$