$7.36 \quad 3 \log _{5} 2+2-x=\log _{5}\left(3^{x}-5^{2-x}\right)$.

$7.37 \quad 25^{\log _{2} \sqrt{x+3}-0.5 \log _{2}\left(x^{2}-9\right)}=\sqrt{2(7-x)}$.
7.36 Here

$3^{x}-5^{2-x}>0$.

Let's use the fact that $2-x=\log _{5} 5^{2-x}$, and write the equation in the form

$\log _{5}\left(2^{3} \cdot 5^{2-x}\right)=\log _{5}\left(3^{x}-5^{2-x}\right)$

Then we have

$2^{3} \cdot 5^{2-x}=3^{x}-5^{2-x} ; 5^{2-x}\left(2^{3}+1\right)=3^{x} ; 5^{2-x}=3^{x-2}$.

Finally, multiplying both sides of the equation by $3^{2-x} \neq 0$, we get

$(5 \cdot 3)^{2-x}=1 \Rightarrow 15^{2-x}=15^{0} \Rightarrow 2-x=0 \Rightarrow x=2$.

This root satisfies the inequality (1).

Answer: $x=2$.