(a)(i) We have the equations:

\(4 = a + \frac{1}{2}b\)

\(3 = a + b\)

Subtract the first equation from the second:

\(3 - 4 = a + b - (a + \frac{1}{2}b)\)

\(-1 = \frac{1}{2}b\)

\(b = -2\)

Substitute \(b = -2\) into \(3 = a + b\):

\(3 = a - 2\)

\(a = 5\)

(a)(ii) \(ff(x) = a + b \sin(a + b \sin x)\)

\(ff(0) = 5 - 2 \sin(5) = 6.92\)

(b) The range of \(g(x) = c + d \sin x\) is \(-4 \leq g(x) \leq 10\).

\(10 = c + d\) and \(-4 = c - d\)

Adding these equations:

\(10 - (-4) = c + d + c - d\)

\(14 = 2c\)

\(c = 7\)

Substitute \(c = 3\) into \(10 = c + d\):

\(10 = 3 + d\)

\(d = 7\)

Alternatively, \(d = -7\) also satisfies the range condition.

(i) The graph of \(y = 2 \cos x\) is a cosine wave with amplitude 2. It completes one full cycle from \(-\pi\) to \(\pi\). The point of intersection with the \(y\)-axis is at \(x = 0\), giving \(y = 2 \cos(0) = 2\). Thus, the coordinates are \((0, 2)\).

(ii) The coordinates of \(P\) are \(\left( \frac{\pi}{3}, 2 \cos \left( \frac{\pi}{3} \right) \right) = \left( \frac{\pi}{3}, 1 \right)\) and \(Q\) are \((\pi, 2 \cos(\pi)) = (\pi, -2)\). The distance \(PQ\) is given by:

\(PQ = \sqrt{\left( \pi - \frac{\pi}{3} \right)^2 + (1 - (-2))^2} = \sqrt{\left( \frac{2\pi}{3} \right)^2 + 3^2} = \sqrt{\frac{4\pi^2}{9} + 9}\)

\(PQ = \sqrt{\frac{4\pi^2}{9} + 9} = \sqrt{\frac{4\pi^2 + 81}{9}} = \sqrt{\frac{4\pi^2 + 81}{9}} \approx 3.7\)

(iii) The equation of the line through \(P\) and \(Q\) is:

\(y - 1 = \frac{-2 - 1}{\pi - \frac{\pi}{3}} (x - \frac{\pi}{3})\)

\(y - 1 = \frac{-3}{\frac{2\pi}{3}} (x - \frac{\pi}{3})\)

\(y - 1 = -\frac{9}{2\pi} (x - \frac{\pi}{3})\)

Setting \(y = 0\) to find \(h\):

\(0 - 1 = -\frac{9}{2\pi} (h - \frac{\pi}{3})\)

\(-1 = -\frac{9}{2\pi} h + \frac{3}{2}\)

\(-\frac{5}{2} = -\frac{9}{2\pi} h\)

\(h = \frac{5}{9} \pi\)

Setting \(x = 0\) to find \(k\):

\(k - 1 = -\frac{9}{2\pi} (0 - \frac{\pi}{3})\)

\(k - 1 = \frac{3}{2}\)

\(k = \frac{5}{2}\)

(i) To find the \(x\)-coordinate of \(A\), set \(\tan x = \cos x\).

This implies \(\sin x = \cos^2 x\).

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we have \(\sin x = 1 - \sin^2 x\).

Rearrange to form a quadratic equation: \(\sin^2 x + \sin x - 1 = 0\).

Solving this quadratic equation gives \(\sin x = 0.618\).

Thus, \(x = \sin^{-1}(0.618) = 0.666\).

(ii) For point \(B\), the \(x\)-coordinate is \(\pi - 0.666 = 2.475\).

Calculate the \(y\)-coordinate using \(y = \tan(2.475)\) or \(y = \cos(2.475)\).

This gives \(y = -0.787\).

Therefore, the coordinates of \(B\) are \((2.48, -0.787)\).

(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 \(\theta = \frac{\pi}{2}\).

To find \(c\), we use the fact that the graph crosses the \(x\)-axis at \(C(\cos^{-1} c, 0)\). At this point, \(y = 0\), so:

\(a \, \cos(\cos^{-1} c) - b = 0\)

\(a \, c - b = 0\)

\(c = \frac{b}{a}\)

To find \(d\), we use the fact that the graph crosses the \(y\)-axis at \(D(0, d)\). At this point, \(x = 0\), so:

\(y = a \, \cos(0) - b\)

\(y = a \, (1) - b\)

\(d = a - b\)