Answer
See the explanation below.
Work Step by Step
(a)
Let us consider the provided function $f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\
& \text{12 if }x=\text{4} \\
\end{align} \right.$
To find $f\left( -2 \right)$, put $x=-2$ in function $f\left( x \right)=\frac{{{x}^{2}}-1}{x-1}$.
When $x\ne 1$,
$\begin{align}
& f\left( x \right)=\frac{{{x}^{2}}-1}{x-1} \\
& f\left( -2 \right)=\frac{{{\left( -2 \right)}^{2}}-1}{-2-1} \\
& =\frac{4-1}{-3} \\
& =-1.
\end{align}$
(b)
Let us consider provided function $f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\
& \text{12 if }x=\text{4} \\
\end{align} \right.$.
To find $f\left( 1 \right)$, put $x=1$ in function $f\left( x \right)=12$.
When $x=1$,
$f\left( 1 \right)=12$.
(c)
Let us consider the provided function $f\left( x \right)=\left\{ \begin{align}
& \frac{{{x}^{2}}-1}{x-1}\text{ if }x\ne 1 \\
& \text{12 if }x=\text{4} \\
\end{align} \right.$.
To find $f\left( 2 \right)$, put $x=2$ in function $f\left( x \right)=\frac{{{x}^{2}}-1}{x-1}$.
When $x\ne 1$,
$\begin{align}
& f\left( x \right)=\frac{{{x}^{2}}-1}{x-1} \\
& f\left( 2 \right)=\frac{{{\left( 2 \right)}^{2}}-1}{2-1} \\
& =\frac{4-1}{1} \\
& =3.
\end{align}$
