(a) Given \(\sin^{-1}(3x) = -\frac{1}{3}\pi\), we have \(3x = \sin\left(-\frac{1}{3}\pi\right) = -\frac{\sqrt{3}}{2}\).

Solving for \(x\), we get \(x = \frac{-\sqrt{3}}{2 \times 3} = -\frac{\sqrt{3}}{6}\).

(b) The equation is \(2 \cos \theta \sin \theta - 2 \cos \theta - \sin \theta + 1 = 0\).

Factorise as \((2\cos \theta - 1)(\sin \theta - 1) = 0\).

This gives \(2\cos \theta - 1 = 0\) or \(\sin \theta - 1 = 0\).

For \(2\cos \theta - 1 = 0\), \(\cos \theta = \frac{1}{2}\), so \(\theta = \frac{\pi}{3}\).

For \(\sin \theta - 1 = 0\), \(\sin \theta = 1\), so \(\theta = \frac{\pi}{2}\).