#### Answer

See below:

#### Work Step by Step

(a)
Consider the function:
$f\left( x \right)={{\left( x+2 \right)}^{3}}$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{\left( x+2 \right)}^{3}}$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y={{\left( x+2 \right)}^{3}}$
Step 2: Interchange the variables x and y. So, the equation is:
$x={{\left( y+2 \right)}^{3}}$
Step 3: Solve the equation for y.
The variable y has to be isolated. As $\sqrt[3]{{{y}^{3}}}=y$ , take cube root on both sides of the equation. So, the equation becomes,
${{\left( x \right)}^{{1}/{3}\;}}={{\left( {{\left( y+2 \right)}^{3}} \right)}^{{1}/{3}\;}}$
Simplify the powers. So,
$\sqrt[3]{x}=y+2$
Subtract $2$ from both sides of the equation and simplify.
$\begin{align}
& \sqrt[3]{x}-2=y+2-2 \\
& \sqrt[3]{x}-2=y \\
\end{align}$
The equation is $y=\sqrt[3]{x}-2$. Because the obtained equation defines y as a function of x, then the inverse function exists for the function f.
Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3.
Thus, the inverse function${{f}^{-1}}\left( x \right)$ is obtained as,
${{f}^{-1}}\left( x \right)=\sqrt[3]{x}-2$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}-2$.
(b) Given above.
(c)
The domain and range of the functions in interval notation are:
\[\begin{align}
& \text{domain of }f=\text{range of }{{f}^{-1}}=\left( -\infty ,\infty \right) \\
& \text{range of }f=\text{domain of }{{f}^{-1}}=\left( -\infty ,\infty \right)
\end{align}\]