7.276. $\left\{\begin{array}{l}2^{\sqrt{x y}-2}+4^{\sqrt{x y}-1}=5, \\ \frac{3(x+y)}{x-y}+\frac{5(x-y)}{x+y}=8 .\end{array}\right.$
## Solution.

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

Transform the first equation of the system $2^{2 \sqrt{x y}}+2^{\sqrt{x y}}-20=0$ and, solving it as a quadratic equation in terms of $2^{\sqrt{x y}}$, we get $2^{\sqrt{2 x y}}=-5$ (no solutions), or $2^{\sqrt{x y}}=2^{2} \Rightarrow \sqrt{x y}=2, x y=4, y=\frac{4}{x}$. From the second equation of the system, we get $3(x+y)^{2}+5(x-y)^{2}=8\left(x^{2}-y^{2}\right)$. Since $y=\frac{4}{x}$, we have $3\left(x+\frac{4}{x}\right)^{2}+5\left(x-\frac{4}{x}\right)^{2}=8\left(x^{2}-\frac{16}{x^{2}}\right) \Leftrightarrow x^{2}=16$, from which $x_{1}=-4, x_{2}=4$. Then $y_{1}=-1, y_{2}=1$.

Answer: $(-4 ;-1),(4 ; 1)$.