#### Answer

See the explanation below.

#### Work Step by Step

Consider the given function $f\left( x \right)=c$ and the difference quotient formula $\frac{f\left( x+h \right)-f\left( x \right)}{h}$.
Substitute $x+h$ in place of x to find the value of $f\left( x+h \right)$.
In the given function, there is no term that consists of x, so after substitution, it will result in $f\left( x+h \right)=c$.
So,
$\begin{align}
& f\left( x+h \right)=c \\
& f\left( x \right)=c
\end{align}$
Consider the given functions $f\left( x \right),\ g\left( x \right)\text{, and }h\left( x \right)$
In case the function $h\left( x \right)$ is even:
$h\left( -x \right)=h\left( x \right)$
This implies:
$h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)}$
It can be observed that $h\left( -x \right)=h\left( x \right)$ holds if both functions $f\ \text{ and }\ g$ are either even or odd.
When both functions f and g are even:
$\begin{align}
& h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\
& =\frac{f\left( x \right)}{g\left( x \right)} \\
& =h\left( x \right)
\end{align}$
When both functions f and g are odd,
$\begin{align}
& h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\
& =\frac{-f\left( x \right)}{-g\left( x \right)} \\
& =h\left( x \right)
\end{align}$
Hence, $h\left( x \right)$ is even if both functions $f\ \text{ and }\ g$ are either even or odd.
(b)
Consider the given functions $f\left( x \right),\ g\left( x \right)\text{, and }h\left( x \right)$
In case the function $h\left( x \right)$ is odd:
$h\left( -x \right)=-h\left( x \right)$
This implies:
$h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)}$
Thus, $h\left( -x \right)=-h\left( x \right)$ holds if out of the functions $f\ \text{ and }\ g$ one is even and the other is odd.
Case 1: When f is odd and g is even:
$\begin{align}
& h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\
& =\frac{-f\left( x \right)}{g\left( x \right)} \\
& =-h\left( x \right)
\end{align}$
Case 2: When f is even and g is odd:
$\begin{align}
& h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\
& =\frac{f\left( x \right)}{-g\left( x \right)} \\
& =-h\left( x \right)
\end{align}$
Hence, $h\left( x \right)$ is odd if out of the functions $f\ \text{ and }\ g$ one is even and the other is odd.
Hence, the difference quotient formula always gives zero when $f\left( x \right)=c$.
Hence, the statement makes sense.