7. Let $f(x)$ be a monotonic function defined on the interval $(0,+\infty)$. For any $x>0$, we have
$$
f(x)>-\frac{4}{x}, f\left(f(x)+\frac{4}{x}\right)=3 \text {. }
$$

Then $f(8)=$ $\qquad$
7. $\frac{7}{2}$.

From the problem, we know there exists $x_{0}>0$ such that $f\left(x_{0}\right)=3$.
Since $f(x)$ is a monotonic function on the interval $(0,+\infty)$, such an $x_{0}>0$ is unique.
From $f\left(f\left(x_{0}\right)+\frac{4}{x_{0}}\right)=3$, we get
$x_{0}=f\left(x_{0}\right)+\frac{4}{x_{0}}=3+\frac{4}{x_{0}}$
$\Rightarrow x_{0}=4$ or $x_{0}=-1$ (discard).
Thus, $f(x)=4-\frac{4}{x}, f(8)=4-\frac{4}{8}=\frac{7}{2}$.