7.337

$$
\left\{\begin{array}{l}
x y=a^{2} \\
\lg ^{2} x+\lg ^{2} y=2.5 \lg ^{2} a^{2},
\end{array} a<0\right.
$$
Solution. Taking the logarithm of the first equation:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \operatorname { l g } x + \operatorname { l g } y = 2 \operatorname { l g } ( - a ) } \\
{ \operatorname { l g } ^ { 2 } x + ( 2 \operatorname { l g } ( - a ) - \operatorname { l g } x ) ^ { 2 } = 1 0 \operatorname { l g } ^ { 2 } ( - a ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\lg y=2 \lg (-a)-\lg x, \\
\lg ^{2} x-2 \lg (-a) \cdot \lg x-3 \lg ^{2}(-a)=0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \operatorname { l g } y = 2 \operatorname { l g } ( - a ) - \operatorname { l g } x , } \\
{ \operatorname { l g } x = \operatorname { l g } ( - a ) \pm 2 \operatorname { l g } ( - a ) }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\lg x=3 \lg (-a), \lg y=-\lg (-a) \\
\lg x=-\lg (-a), \lg y=3 \lg (-a)
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x=(-a)^{3}, y=(-a)^{-1}, \\
x=(-a)^{-1}, y=(-a)^{3}
\end{array}\right.
\end{aligned}
$$

Answer: $\left(-a^{3},-\frac{1}{a}\right),\left(-\frac{1}{a},-a^{3}\right)$.

$$
7.338\left\{\begin{array}{l}
x^{\log _{8} y}+y^{\log _{8} x}=4 \\
\log _{4} x-\log _{4} y=1
\end{array}\right.
$$

Solution. $\left\{\begin{array}{l}x^{\log _{8} y}+x^{\log _{8} y}=4, \\ \log _{2} x-\log _{2} y=2\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{\log _{8} y}=2, \\ \log _{2} x-\log _{2} y=2 .\end{array}\right.\right.$

Taking the logarithm of the first equation:

$\left\{\begin{array}{l}\log _{8} y \cdot \log _{2} x=\log _{2} 2, \\ \log _{2} x=\log _{2} y+2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{2}^{2} y+2 \log _{2} y-3=0, \\ \log _{2} x=\log _{2} y+2\end{array} \Leftrightarrow\left\{\begin{array}{l}\log _{2} y=-3, \\ \log _{2} y=1, \\ \log _{2} x=\log _{2} y+2 .\end{array} \Leftrightarrow\right.\right.\right.$ $\Leftrightarrow\left[\begin{array}{l}y=\frac{1}{8}, x=\frac{1}{2}, \\ y=2, x=8 .\end{array}\right.$

Answer: $\left(\frac{1}{2} ; \frac{1}{8}\right),(8 ; 2)$.

$$
7.339\left\{\begin{array}{l}
\left(2^{x+y}\right)^{x^{2}-x y-8}=1 \\
\left(0,37^{x-y}\right)^{x^{2}+x y+2 x-16}=1
\end{array}\right.
$$

Solution.

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 2 ^ { ( x + y ) ( x ^ { 2 } - x y - 8 ) } = 2 ^ { 0 } , } \\
{ 0 , 3 7 ^ { ( x - y ) ( x ^ { 2 } + x y + 2 x - 1 6 ) } = 0 , 3 7 ^ { 0 } }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
(x+y)\left(x^{2}-x y-8\right)=0, \\
(x-y)\left(x^{2}+x y+2 x-16\right)=0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\left[\begin{array}{l}
x=-y, \\
x^{2}-x y-8=0, \\
{\left[\begin{array}{l}
x=y, \\
x^{2}+x y+2 x-16=0,
\end{array}\right.}
\end{array} \Rightarrow\right.\right.
\end{aligned}
$$

1) $\left\{\begin{array}{l}x=-y, \\ x=y\end{array} \Leftrightarrow x=y=0\right.$. .
2) $\left\{\begin{array}{l}x=-y, \\ x^{2}+x y+2 x-16=0\end{array} \Leftrightarrow\left\{\begin{array}{l}y=-8, \\ x=8 .\end{array}\right.\right.$
3) $\left\{\begin{array}{l}x=y, \\ x^{2}-x y-8=0 .\end{array} \Rightarrow \varnothing\right.$.
4) $\left\{\begin{array}{l}x^{2}-x y-8=0, \\ x^{2}+x y+2 x-16=0\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left\{\begin{array}{l}x^{2}-x y-8=0, \\ x^{2}+x-12=0\end{array} \Leftrightarrow\left[\begin{array}{l}x=-4, y=-2, \\ x=3, y=\frac{1}{3} .\end{array}\right.\right.$

Answer: $(0 ; 0),(8 ;-8),(-4 ;-2),\left(3, \frac{1}{3}\right)$.