Answer
$\left( -\infty, -3 \right)\cup\left( 3,\infty \right)$
Work Step by Step
Using the properties of equality, the given expression, $
\left| 3x \right|-8\gt1
,$ is equivalent to
\begin{array}{l}\require{cancel}
\left| 3x \right|\gt1+8
\\\\
\left| 3x \right|\gt9
.\end{array}
Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a\text{ or } x\lt-a$, then the expression, $
\left| 3x \right|\gt9
,$ is equivalent to
\begin{array}{l}\require{cancel}
3x\gt9
\\\\
x\gt\dfrac{9}{3}
\\\\
x\gt3
,\\\\\text{ OR }\\\\
3x\lt-9
\\\\
x\lt-\dfrac{9}{3}
\\\\
x\lt-3
.\end{array}
Hence, the solution set is $
\left( -\infty, -3 \right)\cup\left( 3,\infty \right)
.$