Answer
$\text{the interval }
\left( -\infty, -2 \right) \cup \left( \dfrac{4}{3},\infty \right)
$
Work Step by Step
Since for any nonnegative $c$, $|x|\gt c$ implies $x\gt c$ or $x\lt -c$, then the solution to the given inequality, $
|3x+1|\gt5
,$ is
\begin{array}{l}\require{cancel}
3x+1\gt5
\\\\
3x\gt5-1
\\\\
3x\gt4
\\\\
x\gt\dfrac{4}{3}
\\\\\text{ OR }\\\\
3x+1\lt-5
\\\\
3x\lt-5-1
\\\\
3x\lt-6
\\\\
x\lt-\dfrac{6}{3}
\\\\
x\lt-2
.\end{array}
In interval notation, the solution is $
\text{the interval }
\left( -\infty, -2 \right) \cup \left( \dfrac{4}{3},\infty \right)
.$