#### Answer

$\left( -\infty, -1 \right] \cup \left[ \dfrac{5}{3}, \infty\right]$

#### Work Step by Step

Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a,$ then the given inequality, $
|1-3x|\ge4
,$ is equivalent to
\begin{array}{l}\require{cancel}
1-3x\ge4 \text{ OR } 1-3x\le-4
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
1-3x\ge4
\\\\
-3x\ge4-1
\\\\
-3x\ge3
\\\\
x\le\dfrac{3}{-3}
\\\\
x\le-1
\\\\\text{ OR }\\\\
1-3x\le-4
\\\\
-3x\le-4-1
\\\\
-3x\le-5
\\\\
x\ge\dfrac{-5}{-3}
\\\\
x\ge\dfrac{5}{3}
.\end{array}
Hence, the solution is the interval $
\left( -\infty, -1 \right] \cup \left[ \dfrac{5}{3}, \infty\right]
.$