Answer
$\text{a) }
f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6}
\\\\\text{b) }
f(42)g(42)=\dfrac{1,275}{4}
\\\\\text{c) }
f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Substitute the given functions,
\begin{array}{l}\require{cancel}
f(x)=
\dfrac{2}{3}x+\dfrac{1}{3}
\\g(x)=
\dfrac{1}{4}x+\dfrac{3}{4}
,\end{array}
into the required operations.
$\bf{\text{Solution Details:}}$
a)
\begin{array}{l}\require{cancel}
f(g(x))=f\left( \dfrac{1}{4}x+\dfrac{3}{4} \right)
\\\\
f(g(x))=\dfrac{2}{3}\left( \dfrac{1}{4}x+\dfrac{3}{4} \right)+\dfrac{1}{3}
\\\\
f(g(x))=\dfrac{2}{12}x+\dfrac{6}{12} +\dfrac{1}{3}
\\\\
f(g(x))=\dfrac{1}{6}x+\dfrac{1}{2} +\dfrac{1}{3}
\\\\
f(g(x))=\dfrac{1}{6}x+\dfrac{3}{6} +\dfrac{2}{6}
\\\\
f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6}
\end{array}
b)
\begin{array}{l}\require{cancel}
f(x)g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{3} \right) \left( \dfrac{1}{4}x+\dfrac{3}{4} \right)
\\\\
f(42)g(42)=\left( \dfrac{2}{3}(42)+\dfrac{1}{3} \right) \left( \dfrac{1}{4}(42)+\dfrac{3}{4} \right)
\\\\
f(42)g(42)=\left( \dfrac{84}{3}+\dfrac{1}{3} \right) \left( \dfrac{42}{4}+\dfrac{3}{4} \right)
\\\\
f(42)g(42)=\left( \dfrac{85}{3} \right) \left( \dfrac{45}{4} \right)
\\\\
f(42)g(42)=\left( \dfrac{85}{\cancel{3}} \right) \left( \dfrac{\cancel{3}(15)}{4} \right)
\\\\
f(42)g(42)=\dfrac{1,275}{4}
\end{array}
c)
\begin{array}{l}\require{cancel}
f(x)+g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{3} \right) + \left( \dfrac{1}{4}x+\dfrac{3}{4} \right)
\\\\
f(x)+g(x)=\left( \dfrac{2}{3}x+\dfrac{1}{4}x \right) + \left(\dfrac{3}{4}+\dfrac{1}{3} \right)
\\\\
f(x)+g(x)=\left( \dfrac{8}{12}x+\dfrac{3}{12}x \right) + \left(\dfrac{9}{12}+\dfrac{4}{12} \right)
\\\\
f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12}
\end{array}
Therefore,
\begin{array}{l}\require{cancel}
\text{a) }
f(g(x))=\dfrac{1}{6}x+\dfrac{5}{6}
\\\\\text{b) }
f(42)g(42)=\dfrac{1,275}{4}
\\\\\text{c) }
f(x)+g(x)=\dfrac{11}{12}x+\dfrac{13}{12}
.\end{array}