Answer
$\color{blue}{\bf{(a) \sqrt{ -\dfrac{1}{x} +2} ; (-\infty, 0 )\bigcup[ \dfrac{1}{2} ,\infty) }}$
$\color{blue}{\bf{(b) -\dfrac{1}{ \sqrt{x+2} } ; (-2 ,\infty) }}$
Work Step by Step
We are given the two functions $\bf{f}$ and $\bf{g}$
$\bf{f(x) = \sqrt{x+2} }$ and $\bf{g(x) = -\dfrac{1}{x} }$
${\bf(a)}$ We are asked to find $\bf{ ( f \text{ }\omicron\text{ g} )(x) }$ and its domain.
$( f \text{ }\omicron\text{ g} )(x) = \color{blue}{\bf{ \sqrt{ -\dfrac{1}{x} +2} }}$
$ -\dfrac{1}{x} +2 \geq0$ or $ \sqrt{ -\dfrac{1}{x} +2} $ would be imaginary
$ -\dfrac{1}{x} \geq-2$
$ -1 \geq-2x$
$\dfrac{1}{2}\leq{x}$
$ x \neq0$ because division by zero is undefined, and can be any negative number because $-\dfrac{1}{x}$ would be positive so $x< 0$
so the domain is:
$\color{blue}{\bf(-\infty, 0 )\bigcup[ \dfrac{1}{2} ,\infty) }$
${\bf(b)}$ We are asked to find $\bf{ ( g \text{ }\omicron\text{ f} )(x) }$ and its domain.
$( g \text{ }\omicron\text{ f} )(x) =
\color{blue}{\bf{ -\dfrac{1}{ \sqrt{x+2} } }}$
$x+2\geq0$ or $\sqrt{x+2}$ would be imaginary
$x\geq-2$
$ x+2 \neq0$ because $\sqrt{0}=0$ and division by zero is undefined
$ x \neq-2$
$\color{blue}{\bf(-2 ,\infty) }$