(a) We start with the equation \(\sin^{-1}(x^2 - 1) = \frac{1}{3}\pi\).

This implies \(x^2 - 1 = \sin\left(\frac{\pi}{3}\right)\).

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

Solving for \(x^2\), we get \(x^2 = 1 + \frac{\sqrt{3}}{2}\).

Thus, \(x = \pm \sqrt{1 + \frac{\sqrt{3}}{2}}\).

Calculating this gives \(x \approx \pm 1.366\) to 3 decimal places.

(b) We have \(\sin(2\theta + \frac{1}{3}\pi) = \frac{1}{2}\).

The general solution for \(\sin A = \frac{1}{2}\) is \(A = \frac{\pi}{6} + 2k\pi\) or \(A = \frac{5\pi}{6} + 2k\pi\), where \(k\) is an integer.

Thus, \(2\theta + \frac{1}{3}\pi = \frac{\pi}{6}\) or \(2\theta + \frac{1}{3}\pi = \frac{5\pi}{6}\).

Solving these equations:

1. \(2\theta + \frac{1}{3}\pi = \frac{\pi}{6}\)

\(2\theta = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}\)

\(\theta = -\frac{\pi}{12}\) (not in the range \(0 \leq \theta \leq \pi\))

2. \(2\theta + \frac{1}{3}\pi = \frac{5\pi}{6}\)

\(2\theta = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\)

\(\theta = \frac{\pi}{4}\)

3. \(2\theta + \frac{1}{3}\pi = \frac{13\pi}{6}\)

\(2\theta = \frac{13\pi}{6} - \frac{\pi}{3} = \frac{13\pi}{6} - \frac{2\pi}{6} = \frac{11\pi}{6}\)

\(\theta = \frac{11\pi}{12}\)

Thus, the solutions are \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{11\pi}{12}\).