Solution. (a). A “greater than” absolute value gives two separate pieces: \[3x-2<-2 \quad \text {or}\quad 3x-2>2,\] that is \(3x<0\) or \(3x>4\), so \[x<0 \quad \text {or}\quad x>\frac {4}{3}, \quad \text {i.e. } (-\infty ,0)\cup \left (\tfrac {4}{3},\infty \right ).\]

(b). Never cross-multiply: both denominators change sign. Move everything to one side, \[\frac {2}{x-3}-\frac {3}{2x-1}<0 \implies \frac {2(2x-1)-3(x-3)}{(x-3)(2x-1)}<0 \implies \frac {x+7}{(x-3)(2x-1)}<0 .\] The critical values are \(x=-7\), \(x=\frac {1}{2}\) and \(x=3\). Testing the sign of the whole expression on each interval between them:

The expression is negative on \((-\infty ,-7)\) and on \(\left (\frac {1}{2},3\right )\), and all three critical values are excluded — \(-7\) because the inequality is strict, the other two because the expression is undefined there. Hence \[(-\infty ,-7)\cup \left (\tfrac {1}{2},3\right ).\]