Answer: \(x\lt-\frac52\) or \(x\gt3\).

Consider the two possible cases for the modulus.

If

\(3x+2\gt 8+x,\)

then

\(2x\gt6,\)

so

\(x\gt3.\)

If

\(-(3x+2)\gt 8+x,\)

then

\(-3x-2\gt 8+x.\)

Hence

\(-4x\gt10.\)

Dividing by a negative number reverses the inequality, so

\(x\lt-\frac52.\)

Therefore the solution is

\(x\lt-\frac52\quad\text{or}\quad x\gt3.\)