7.262. $\left\{\begin{array}{l}\log _{2}(x-y)=5-\log _{2}(x+y) \\ \frac{\lg x-\lg 4}{\lg y-\lg 3}=-1 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}x-y>0, \\ x+y>0, \\ x>0, \\ y>0, \\ y \neq 3 .\end{array}\right.$

From the condition we have

$$
\left\{\begin{array} { l } 
{ \operatorname { log } _ { 2 } ( x - y ) = \operatorname { log } _ { 2 } \frac { 3 2 } { x + y } , } \\
{ \operatorname { lg } \frac { x } { 4 } = \operatorname { lg } \frac { 3 } { y } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x-y=\frac{32}{x+y}, \\
\frac{x}{4}=\frac{3}{y}
\end{array} \Rightarrow x=\frac{12}{y}\right.\right.
$$

From the first equation we get

$$
\frac{12}{y}-y=\frac{32}{\frac{12}{y}+y} \Leftrightarrow y^{4}+32 y^{2}-144=0, y_{1}=-2, y_{2}=2
$$

Then $x_{1}=-6, x_{2}=6$.

$\left\{\begin{array}{l}y_{1}=-2, \\ x_{1}=-6\end{array}\right.$ do not satisfy the domain of definition.

Answer: $(6 ; 2)$.