(i) (a) Since \(\theta\) is a reflex angle, it is in the 4th quadrant where sine is negative. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - k^2\)

\(\sin \theta = -\sqrt{1-k^2}\)

(i) (b) \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{1-k^2}}{k}\)

(ii) Since \(\theta\) is in the 4th quadrant, \(2\theta\) will be in the range of 540° to 720°, which corresponds to the 3rd and 4th quadrants. In both these quadrants, \(\sin 2\theta\) is negative.

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

(i) Since \(\cos x = p\) and \(x\) is an acute angle, we use the identity \(\sin^2 x + \cos^2 x = 1\). Therefore, \(\sin^2 x = 1 - \cos^2 x = 1 - p^2\). Taking the positive square root (since \(x\) is acute), \(\sin x = \sqrt{1 - p^2}\).

(ii) \(\tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{1 - p^2}}{p}\).

(iii) Using the identity \(\tan(90^\circ - x) = \cot x\), we have \(\tan(90^\circ - x) = \frac{1}{\tan x} = \frac{p}{\sqrt{1 - p^2}}\).

(i) We have \(a = \\sin \theta - 3 \\cos \theta\) and \(b = 3 \\sin \theta + \\cos \theta\).

Calculate \(a^2 + b^2\):

\(a^2 = (\\sin \theta - 3 \\cos \theta)^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta\)

\(b^2 = (3 \\sin \theta + \\cos \theta)^2 = 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

Adding these, \(a^2 + b^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta + 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

\(= 10 \\sin^2 \theta + 10 \\cos^2 \theta\)

Using \(\\sin^2 \theta + \\cos^2 \theta = 1\), we get \(a^2 + b^2 = 10\).

(ii) Given \(2a = b\), substitute \(a\) and \(b\):

\(2(\\sin \theta - 3 \\cos \theta) = 3 \\sin \theta + \\cos \theta\)

\(2 \\sin \theta - 6 \\cos \theta = 3 \\sin \theta + \\cos \theta\)

\(-\\sin \theta = 7 \\cos \theta\)

\(\\tan \theta = -\frac{1}{7}\)

\(\theta = \\arctan(-\frac{1}{7}) \approx 98.1^\circ\) and \(\theta = 98.1^\circ + 180^\circ = 278.1^\circ\).

(i) We have \(gf(x) = 1\), which means \(g(f(x)) = 1\). Therefore, \(\cos\left(\frac{1}{2}x + \frac{\pi}{6}\right) = 1\).

The cosine function equals 1 at \(0\), so \(\frac{1}{2}x + \frac{\pi}{6} = 0\).

Solving for \(x\), we get:

\(\frac{1}{2}x = -\frac{\pi}{6}\)

\(x = -\frac{\pi}{3}\).

(ii) We have \(fg(x) = 1\), which means \(f(g(x)) = 1\). Therefore, \(\frac{1}{2}\cos x + \frac{\pi}{6} = 1\).

Solving for \(\cos x\), we get:

\(\frac{1}{2}\cos x = 1 - \frac{\pi}{6}\)

\(\cos x = 2\left(1 - \frac{\pi}{6}\right)\)

\(\cos x = 0.9528\)

Solving for \(x\), we find \(x = \pm 0.31\) (to 2 decimal places).