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