\(\sqrt{x-8}\) is undefined if the argument \(x-8\) is negative, so you first need to require that
\(x-8\ge0\implies x\ge8\)
We're not done yet, though, because \(\dfrac1{\sqrt{x-8}}\) still doesn't exist when \(x=8\), so we remove this from the domain and we're left with \(x>8\), or in interval notation, \((8,\infty)\)
To find the range, consider the limits of the function as you approach either endpoint of the domain.
\(\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty\)
\(\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0\)
Since \(\dfrac1{\sqrt{x-8}}\) is positive everywhere, the range is \((0,\infty)\)
