7. Let $f(x)$ be a monotonic function defined on $(0,+\infty)$, for any $x>0$ we have $f(x)>-\frac{4}{x}, f\left(f(x)+\frac{4}{x}\right)=3$, then $f(8)=$ $\qquad$ .
Solve
$f\left(f(x)+\frac{4}{x}\right)=3 \Rightarrow f(x)+\frac{4}{x}=c>0 \Rightarrow f(x)=c-\frac{4}{x}$, substitute $f(c)=3$ to get $c-\frac{4}{c}=3$ $\Rightarrow c^{2}-3 c-4=0 \Rightarrow c=4(c=-1$ is discarded $)$. Therefore, $f(x)=4-\frac{4}{x} \Rightarrow f(8)=4-\frac{4}{8}=4-\frac{1}{2}=\frac{7}{2}$.