7.277. $\left\{\begin{array}{l}x^{y}=2, \\ (2 x)^{y^{2}}=64(x>0)\end{array}\right.$
## Solution.

Domain of definition: $0<x \neq 1$.

Taking the logarithm of both equations of the system to the base 2, we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \operatorname { log } _ { 2 } x ^ { y } = \operatorname { log } _ { 2 } 2 , } \\
{ \operatorname { log } _ { 2 } ( 2 x ) ^ { y ^ { 2 } } = \operatorname { log } _ { 2 } 6 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y \log _{2} x=1, \\
y^{2}\left(1+\log _{2} x\right)=6
\end{array} \Rightarrow \log _{2} x=\frac{1}{y}\right.\right. \\
& y^{2}\left(1+\frac{1}{y}\right)=6, y^{2}+y-6=0, y_{1}=-3, y_{2}=2
\end{aligned}
$$

Then $\log _{2} x=\frac{1}{-3}, x_{1}=\frac{1}{\sqrt[3]{2}}$, or $\log _{2} x=\frac{1}{2}, x_{2}=\sqrt{2}$.

Answer: $\left(\frac{\sqrt[3]{4}}{2} ;-3\right),(\sqrt{2} ; 2)$.