Answer
a) $h(32)=-9$
b) $x =15$
c) $\text{Domain: all real numbers}
\\\text{Range: all real numbers}
$
Work Step by Step
a) To find $h(32),$ substitute $x$ with $32$ in the given function, $h(x)=-\dfrac{5}{8}x+11.$ Hence,
\begin{array}{l}\require{cancel}
h(x)=-\dfrac{5}{8}x+11
\\\\
h(32)=-\dfrac{5}{8}(32)+11
\\\\
h(32)=-20+11
\\\\
h(32)=-9
.\end{array}
b) Replace $h(x)$ with $\dfrac{13}{8}$ in the given function, $h(x)=-\dfrac{5}{8}x+11.$ Hence,
\begin{array}{l}\require{cancel}
h(x)=-\dfrac{5}{8}x+11
\\\\
\dfrac{13}{8}=-\dfrac{5}{8}x+11
\\\\
8\left( \dfrac{13}{8} \right) =8\left( -\dfrac{5}{8}x+11 \right)
\\\\
13 =-5x+88
\\\\
5x =88-13
\\\\
5x =75
\\\\
x =\dfrac{75}{5}
\\\\
x =15
.\end{array}
c) The domain and the range of linear functions in the form $f(x)=mx+b$ is the set of all real numbers. Hence, the given function has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Domain: all real numbers}
\\\text{Range: all real numbers}
.\end{array}