7.278.

$$
\left\{\begin{array}{l}
\frac{x^{2}}{y}+\frac{y^{2}}{x}=12 \\
2^{-\log _{2} x}+5^{\log _{s} \frac{1}{y}}=\frac{1}{3}
\end{array}\right.
$$
Solution.

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

Transform the first equation of the system

$x^{3}+y^{3}=12 x y \Leftrightarrow(x+y)\left((x+y)^{2}-3 x y\right)=12 x y$.

The second equation, using the equality $a^{\log _{a} b}=b$, can be represented as $\frac{1}{x}+\frac{1}{y}=\frac{1}{3} \Leftrightarrow 3(x+y)=x y$. The system has the form $\left\{\begin{array}{l}(x+y)\left((x+y)^{2}-3 x y\right)=12 x y, \\ 3(x+y)=x y .\end{array}\right.$ If $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$ then $\left\{\begin{array}{l}u\left(u^{2}-3 v\right)=12 v, \\ 3 u=v\end{array} \Rightarrow v=3 u, u\left(u^{2}-3(3 u)\right)=12 \cdot 3 u \Leftrightarrow\right.$ $\Leftrightarrow u^{3}-9 u^{2}-36 u=0 \Leftrightarrow u\left(u^{2}-9 u-36\right)=0$, from which $u_{1}=0, u_{2}=-3$, $u_{2}=12$. Then $v_{1}=0, v_{2}=-9, v_{3}=36$. The original system of equations is equivalent to three systems:

1) $\left\{\begin{array}{l}x+y=0, \\ x y=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x=0, \\ y=0\end{array}\right.\right.$ does not satisfy the domain of definition;
2) $\left\{\begin{array}{l}x+y=-3, \\ x y=-9,\end{array}\right.$ here $x y<0$, which does not satisfy the domain of definition;
3) $\left\{\begin{array}{l}x+y=12, \\ x y=36\end{array} \Leftrightarrow\left\{\begin{array}{l}x=6 \\ y=6\end{array}\right.\right.$

Answer: $(6 ; 6)$.