7.142. $\left\{\begin{array}{l}\log _{2} x+\log _{4} y=4, \\ \log _{4} x+\log _{2} y=5\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\log _{2} x+\log _{4} y=4, \\ 
\log _{4} x+\log _{2} y=5
\end{array}\right.
\]
Solution.

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

Let's switch to base 2. We have

$\left\{\begin{array}{l}\log _{2} x+\frac{1}{2} \log _{2} y=4, \\ \frac{1}{2} \log _{2} x+\log _{2} y=5\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \log _{2} x+\log _{2} y=8, \\ \log _{2} x+2 \log _{2} y=10\end{array} \Rightarrow\right.\right.$

$\Rightarrow\left\{\begin{array}{l}\log _{2} x^{2} y=8, \\ \log _{2} x y^{2}=10\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2} y=2^{8}, \\ x y^{2}=2^{10} .\end{array}\right.\right.$

From the first equation of the system $y=\frac{2^{8}}{x^{2}}$. From the second equation $x \cdot\left(\frac{2^{8}}{x^{2}}\right)^{2}=2^{10}, x^{3}=2^{6}$, hence $x=4, y=16$.

Answer: $(4 ; 16)$