Answer
$a.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
$\ \ \ \ \quad g\circ f=\frac{1}{\sqrt{x+1}}$
$b.\quad \mathrm{Domains:}$
$i.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
$\mathrm{Domain:}\ \ (-\infty,-1]\cup (0,\infty)$
$ii.\quad g\circ f=\frac{1}{\sqrt{x+1}}$
$\mathrm{Domain:}\ \ (-1,\infty)$
$c.\quad \mathrm{Ranges:}$
$i.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
$\mathrm{Range:}\ \ [0,1)\cup (1,\infty)$
$ii.\quad g\circ f=\frac{1}{\sqrt{x+1}}$
$\mathrm{Range:}\ \ (0,\infty)$
Work Step by Step
$a.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
$\ \ \ \ \quad g\circ f=\frac{1}{\sqrt{x+1}}$
$b.\quad \mathrm{Domains:}$
$i.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
We can take square root only of positive numbers.
$\frac{1}{x}+1\ge 0\quad \Rightarrow \quad \frac{1}{x}\ge -1$
It is true for any positive value of $\ x\ $, where it cannot be equal to zero becauase of it's position in the denominator. When $\ x\ $ is negative, then we will have:
$\frac{1}{x}\ge -1\ \ \rightarrow \ \ 1\le -x\ \ \rightarrow \ \ x\le -1$
$\mathrm{Domain:}\ \ (-\infty,-1]\cup (0,\infty)$
$ii.\quad g\circ f=\frac{1}{\sqrt{x+1}}$
$x+1\ge0\ $ where $\ x\ge-1\ $, provided that $\ x\ $ cannot be $\ -1.$ This is because when $\ x=-1\ $, we will get zero in the denominator which is not accepted.
$\mathrm{Domain:}\ \ (-1,\infty)$
$c.\quad \mathrm{Ranges:}$
$i.\quad f\circ g=\sqrt{\frac{1}{x}+1}$
The square root will always yield positive number as a result. The function can get any positive number in result except $\ 1\ $, because the fraction $\ \frac{1}{x}\ $ must be zero in order to get $\ 1\ $ as the result.
$\mathrm{Range:}\ \ [0,1)\cup (1,\infty)$
$ii.\quad g\circ f=\frac{1}{\sqrt{x+1}}$
Fraction with positive denominator can obtain any positive value.
$\mathrm{Range:}\ \ (0,\infty)$
