Split the modulus into two cases.

\(4x-3=7\) or \(4x-3=-7\).

From \(4x-3=7\), we get \(4x=10\), so \(x=\tfrac52\).

From \(4x-3=-7\), we get \(4x=-4\), so \(x=-1\).

So the solutions are \(x=-1\) and \(x=\tfrac52\).

Split into two cases.

\(1-2x=5\) or \(1-2x=-5\).

From \(1-2x=5\), we get \(-2x=4\), so \(x=-2\).

From \(1-2x=-5\), we get \(-2x=-6\), so \(x=3\).

So the solutions are \(x=-2\) and \(x=3\).

Split into two cases.

\(\dfrac{3x-2}{5}=4\) or \(\dfrac{3x-2}{5}=-4\).

From the first, \(3x-2=20\), so \(3x=22\) and \(x=\tfrac{22}{3}\).

From the second, \(3x-2=-20\), so \(3x=-18\) and \(x=-6\).

So the solutions are \(x=-6\) and \(x=\tfrac{22}{3}\).

Split into two cases.

\(\dfrac{x}{3}+2=3\) or \(\dfrac{x}{3}+2=-3\).

From the first, \(\dfrac{x}{3}=1\), so \(x=3\).

From the second, \(\dfrac{x}{3}=-5\), so \(x=-15\).

So the solutions are \(x=-15\) and \(x=3\).

Split into two cases.

\(\dfrac{x+2}{3}-\dfrac{2x}{5}=2\) or \(\dfrac{x+2}{3}-\dfrac{2x}{5}=-2\).

Multiply through by \(15\).

First case: \(5(x+2)-6x=30\), so \(-x+10=30\), hence \(x=-20\).

Second case: \(5(x+2)-6x=-30\), so \(-x+10=-30\), hence \(x=40\).

So the solutions are \(x=-20\) and \(x=40\).