Given f(x) = (x)/(x-3).

The domain of $f^{-1}(x)$ is restricted when the denominator is zero.

The domain of $f^{-1}(x)$ is $\{x \in \mathbb{R} \mid x \neq 1\}$.
The inverse function is $\boxed{f^{-1}(x) = \frac{3x}{x-1}}$ with domain $\boxed{\{x \in \mathbb{R} \mid x \neq 1\}}$.

For $g(x) = \sqrt{x+2}$ to be defined, $x+2 \ge 0 \implies x \ge -2$.
For $f(g(x))$ to be defined, the denominator $\sqrt{x+2}-3$ cannot be zero.

Combining these conditions, the domain of $fg(x)$ is $\{x \in \mathbb{R} \mid x \ge -2 \text{ and } x \neq 7\}$.

Let $u = g(x) = \sqrt{x+2}$.
From the domain of $fg(x)$, we know $x \ge -2$ and $x \neq 7$.
This implies $u = \sqrt{x+2} \ge \sqrt{-2+2} = 0$.
Also, $u = \sqrt{x+2} \neq \sqrt{7+2} = \sqrt{9} = 3$.
So, the range of $u$ is $[0, \infty) \setminus \{3\}$.
Now we need to find the range of $f(u) = \frac{u}{u-3}$ for $u \in [0, \infty) \setminus \{3\}$.
Let $y = \frac{u}{u-3}$.
To find the range, we can express $u$ in terms of $y$: