7. Let the function $f(x)$ defined on $(0,+\infty)$ satisfy that for any $x \in(0,+\infty)$, we have $f(x)>-\frac{3}{x}$, and $f\left(f(x)+\frac{3}{x}\right)=2$. Then $f(5)=$ $\qquad$ .
7. $\frac{12}{5}$.

From the fact that $f(x)$ is monotonically increasing on $(0,+\infty)$ and $f(x)+\frac{3}{x}>0$, we know that if $f(x)=2$, then $2+\frac{3}{x}=x$, solving this gives $x=3$ (discard the negative value), thus $f(x)=3-\frac{3}{x}$. Substituting $x=5$, we get $f(5)=\frac{12}{5}$.