7.136. $\left\{\begin{array}{l}3^{2 \sqrt{x}-\sqrt{y}}=81 \\ \lg \sqrt{x y}=1+\lg 3\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
3^{2 \sqrt{x}-\sqrt{y}}=81 \\
\lg \sqrt{x y}=1+\lg 3
\end{array}\right.
\]
## Solution.

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

From the first equation of the system, we have $3^{2 \sqrt{x}-\sqrt{y}}=3^{4}, 2 \sqrt{x}-\sqrt{y}=4$, $\sqrt{y}=2 \sqrt{x}-4$. From the second equation of the system, we get $\sqrt{x y}=30$, $\sqrt{x} \cdot \sqrt{y}=30$. The system takes the form

$$
\left\{\begin{array}{l}
\sqrt{y}=2 \sqrt{x}-4 \\
\sqrt{x} \cdot \sqrt{y}=30
\end{array} \Rightarrow(\sqrt{x})^{2}-2 \sqrt{x}-15=0\right.
$$

from which $\sqrt{x}=5$ or $\sqrt{x}=-3$ (not suitable). Then $\sqrt{y}=6$. Therefore, $x=25, y=36$.

Answer: $\quad(25 ; 36)$