32. $\left\{\begin{array}{l}\log _{y} \log _{y} x=\log _{x} \log _{x} y \\ \log _{a}^{2} x+\log _{a}^{2} y=8 .\end{array}\right.$
32. We will transform the left and right parts of the first equation to logarithms with base $a$. We get $\left(\log _{a} \log _{a} x-\log _{a} \log _{a} y\right)\left(\frac{1}{\log _{a} y}+\frac{1}{\log _{a} x}\right)=0 . \quad$ This system is equivalent to the following two:
a) $\left\{\begin{array}{l}\log _{a} \log _{a} x-\log _{a} \log _{a} y=0 . \\ \log _{a}^{2} x+\log _{a}^{2} y=8\end{array}\right.$
b) $\left\{\begin{array}{l}\frac{1}{\log _{a} y}+\frac{1}{\log _{a} x}=0 \\ \log _{a}^{2} x+\log _{a}^{2} y=8\end{array}\right.$
Answer:

$$
x_{1}=y_{1}=a^{2}
$$

$$
x_{2}=y_{2}=a^{-2}
$$