You must also ensure that $\sqrt{25-x^2} \neq 0$ otherwise you would be dividing by zero. Therefore you must have $25 \neq x^2$, which means the domain is $]-5,5[$.
To find the range, notice that the image of an interval by a continuous function is always an interval by the Intermediate Value Theorem (you might not have seen that, but let's not worry about it). The point is you must find the maximum and the minimum value of $f$, and then see if they are attained.
The minimum value of $f(x) = \frac 1{g(x)}$ is attained at the maximum value of $g(x)$ in general. Therefore, since the maximum value of $\sqrt{25-x^2}$ is attained at $x=0$, the minimum of $f$ is $\frac 15$.
There is no maximum for $f$, because $\sqrt{25 - x^2}$ can get arbitrarily close to $0$, hence $f$ can get as big as you want. The range is therefore $[1/5,\infty[$.
Hope that helps,
NOTE: I'm going to use the more precise term "image" instead of "range" because I've heard people use range to mean either image OR codomain and I'd prefer to be unambiguous.
Given two functions $f: X \to Y$ and $g: Y \to Z$, the composition $g\circ f$ has domain $X$, codomain $Z$, and the image (range) is given by the image of the restriction of $g$ to the image of $f$.
Example: Let $f: \Bbb R \to \Bbb R$ be given by $f(x)=x^2$ and let $g: \Bbb R \to \Bbb R$ be given by $g(x) = 2x+3$. Then the composition $g\circ f$ is given by $(g\circ f)(x)=g(f(x)) = g(x^2) = 2x^2+3$. Because the domain of $f$ is $\Bbb R$, the domain of $g\circ f$ is $\Bbb R$. Because the codomain of $g$ is $\Bbb R$, the codomain of $g\circ f$ is $\Bbb R$. With that we can write down the signature of the function as $g\circ f: \Bbb R \to \Bbb R$.
But the image of the function takes a little thought to figure out. We need to find the image of the restriction of $g$ to the image of $f$. So first, what's the image of $f$? We can see that it is the nonnegative real numbers $\Bbb R_{\ge 0}$. So then what is the image of the function $\left.g\right|_{\operatorname{Im}(f)}: \Bbb R_{\ge 0} \to \Bbb R$? I'm not sure how to show it without some calculus, but hopefully it's pretty clear that it's the interval $[3, \infty)$. So that is the image of $g\circ f$.
So, to sum up: $$\begin{align}\operatorname{Dom}(g\circ f) &= \Bbb R \\ \operatorname{Codom}(g\circ f) &= \Bbb R \\ \operatorname{Im}(g\circ f) &= [3,\infty)\end{align}$$
As to your question about notation, I haven't seen that notation before (for whatever that's worth), but as long as you define it somewhere on whatever paper/ homework you're using it in, it should be fine.
Best Answer
suppose $x=n+p\\0\leq p<1$ $$\quad{f(x)=\sqrt{4[x]-[4x]}=\\\sqrt{4[n+p]-[4(n+p)]}=\\ \sqrt{4n-[4n+4p]}=\\ \sqrt{4n-4n-[4p]}=\\ \sqrt{-[4p]}\\\implies-[4p]\geq 0\\ [4p]\leq 0 \implies [4p]=0 \\0 \leq 4p<1 \\0\leq p <\frac{1}{4} \to\\D_{f(x)}=[n,n+\frac{1}{4}) ,n \in \mathbb{Z}\\R_{f(x)}=\{\sqrt{0}\}=\{0\} }$$