Solution: $\left(-1,-\frac{\sqrt{2}}{2}\right] \cup\{0\}$.
$f(x)$ is an odd function on $\mathbb{R}$, and from $f(x+1)+f(x)=0$, we have $f(x+2)=-f(x+1)=f(x)$. Thus, $f(x)$ has a period of 2.
Therefore, $f(1)=f(-1+2)=f(-1)=-f(1)$, so $f(1)=0$.
Thus, $f(-3)=f(-3+4)=f(1)=0$.
When $x \in\left[-3,-\frac{5}{2}\right]$, $x+2 \in\left[-1,-\frac{1}{2}\right]$, and $-x-2 \in\left[\frac{1}{2}, 1\right]$.
Hence, $f(x)=f(x+2)=-f(-x-2)=-\sqrt{-x-2}$.
Therefore, when $x \in\left[-3,-\frac{5}{2}\right]$, the range of $f(x)$ is $\left(-1,-\frac{\sqrt{2}}{2}\right] \cup\{0\}$.
5. Given a function $f(x)$ defined on $\mathbb{R}$, for any $x \in \mathbb{R}$, we have
$$
\begin{array}{l}
f(-x)+f(x)=0 \\
f(x+1)+f(x)=0
\end{array}
$$

and when $x \in(0,1)$, $f(x)=\sqrt{x}$. Then, when $x \in\left[-3,-\frac{5}{2}\right]$, the range of $f(x)$ is