To find the range you want, namely of $g(f(x))=F(x),$ note that $F(x)$ is never negative. Also, it is defined and continuous at all real $x.$ Then it is an even function of $x.$ Thus, it suffices to consider only the range for positive $x,$ say.
Note that $F(0)>0.$ Also, as $x\to +\infty,$ we have $F(x)\to \sqrt 3.$ Thus, the range is at least $[F(0),\sqrt 3),$ by IVT. It only remains to see if the function attains $\sqrt 3$ at any point, or if it ever falls below $F(0).$ The first is easily answered in the negative (set $F(x)=\sqrt 3$ to deduce a contradiction), so the range is half-open. So does the function ever go below $e^2-4e+3$? In particular we may ask whether the function ever attains the minimum possible here, namely $0.$ Thus, setting $F(x)=0,$ we see that we want to see if there are real solutions to the quadratic equation $p^2-4p+3=0,$ where $p=e^{-x^2+1}.$ This has solutions. In particular, we get that $e^{-x^2+1}=1,$ or in other words that $$-x^2+1=0.$$ Thus $F(x)$ vanishes in at least two points.
It therefore is the case that the range is $[0,\sqrt 3).$
We must not take roots of negatives and we must not divide by zero, hence the maximal domain of $f$in $\Bbb R$ is $[-2,3)\cup(3,\infty]$.
In its domain, $f$ is obviously continuous (as composition of continuous functions $+,-,\cdot,/,\sqrt{}$).
We have $f(x)\to 0$ as $x\to\infty$ and $f(x)\to +\infty$ as $x\to 3^+$. Then by the intermediate value theorem, the range of $f$ contains $(0,\infty)$.
We have $f(x)\le 0$ for all $x\in [-2,3)$ and $f(-2)=0$ and $f(x)\to-\infty $ as $x\to 3^-$. Again by the intermediate value theorem, the range of $f$ contains $(-\infty,0]$. Together, these two findings show that the range of $f$ is $\Bbb R$.
If you want, you can avoid limits in the arguments of the preceding paragraph (though using them clarifies much better what is behind the result):
For $x>3$, we have
$$ f(x)=\frac{\sqrt{x+2}}{x^2-9}<\frac{\sqrt{x+3}}{x^2-9}<\frac{x+3}{x^2-9}=\frac1{x-3}$$
and
$$f(x)=\frac{\sqrt{x+2}}{x^2-9}>\frac2{x^2-9}. $$
Assume $y>0$. Let $a=\frac1y+3>3$ and $b=\sqrt{\frac2y+9}>3$. Then $f(a)>y>f(b)$ and by continuity of $f$ on $[a,b]\subset(3,\infty)$, we can apply the inermediate value theorem to deduce the existence of $c\in (a,b)$ with $f(c)=y$.
You can find similar nice (i.e., explicitly invertible) bounds for $2<x<3$ and uses these to show similarly that all negative $y$ are in the range. And of course $f(-2)=0$ shos that also $0$ is in the range.
Best Answer
HINT: Note that
$$f(x)^2=(x-1)+2\sqrt{(x-1)(5-x)}+(5-x)=4+2\sqrt{(x-1)(5-x)}\;.$$
Clearly $2\sqrt{(x-1)(5-x)}\ge 0$ for all $x\in[1,5]$, so $4$ is the minimum value of $f(x)^2$, and $2=f(1)=f(5)$ is the minimum value of $f(x)$. To find the maximum value, you must find the maximum value of $(x-1)(5-x)$. The graph of $y=(x-1)(5-x)$ is a parabola opening down; where is its vertex?