8.403. $\left\{\begin{array}{l}\sin x \sin y=0.75, \\ \tan x \tan y=3 .\end{array}\right.$
Solution.

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

Rewrite the second equation of the system as

$$
\frac{\sin x \sin y}{\cos x \cos y}=3 \Leftrightarrow \frac{0.75}{\cos x \cos y}=3, \cos x \cos y=0.25
$$

Then the given system has the form

$\left\{\begin{array}{l}\sin x \sin y=0.75 \\ \cos x \cos y=0.25 .\end{array}\right.$

By adding the second equation of this system to the first, and then subtracting the first equation from the second, we get

$$
\left\{\begin{array} { l } 
{ \operatorname { cos } x \operatorname { cos } y + \operatorname { sin } x \operatorname { sin } y = 1 , } \\
{ \operatorname { cos } x \operatorname { cos } y - \operatorname { sin } x \operatorname { sin } y = - \frac { 1 } { 2 } , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\cos (x-y)=1 \\
\cos (x+y)=-\frac{1}{2}
\end{array}\right.\right.
$$

$$
\left\{\begin{array}{l}
x-y=2 \pi k_{1} \\
x+y= \pm \frac{2}{3} \pi+2 \pi k_{2},
\end{array} k_{1} \text { and } k_{2} \in Z . \Leftrightarrow\left\{\begin{array}{l}
x= \pm \frac{\pi}{3}+\pi\left(k_{1}+k_{2}\right), \\
y= \pm \frac{\pi}{3}+\pi\left(k_{2}-k_{1}\right)
\end{array} k_{1} \text { and } k_{2} \in Z\right.\right.
$$

Answer: $x= \pm \frac{\pi}{3}+\pi\left(k_{1}+k_{2}\right), y= \pm \frac{\pi}{3}+\pi\left(k_{2}-k_{1}\right), k_{1}$ and $k_{2} \in Z$.