(a) To sketch the graph of \(y = f^{-1}(x)\), reflect the graph of \(y = f(x)\) across the line \(y = x\). Ensure the line \(y = x\) is drawn correctly and the inverse graph reaches \(y = 2\pi\) and \(x\)-axis, crossing the line \(y = x\) in the correct squares.

(b) Start with \(y = 3 + 2 \sin \frac{1}{4}x\). Rearrange to find \(\sin \frac{1}{4}x = \frac{y-3}{2}\). Solve for \(x\) to get \(x = 4 \sin^{-1} \left( \frac{y-3}{2} \right)\). Thus, \(f^{-1}(x) = 4 \sin^{-1} \left( \frac{x-3}{2} \right)\).

(c) Extend the graph of \(g(x)\) from \(y\)-intercept to \(-2\pi\). The graph should start to level off as \(x \to -2\pi\). The function \(g(x)\) has an inverse because it is always increasing, passing the horizontal line test.

(i) (a) For \(k = 2\), \(f(x) = 3 - 4 \cos^2 x\). The maximum value of \(\cos^2 x\) is 1, and the minimum is 0. Thus, the range of \(f(x)\) is from \(3 - 4(1) = -1\) to \(3 - 4(0) = 3\). Therefore, the range is \([-1, 3]\).

(b) To solve \(f(x) = 1\), set \(3 - 4 \cos^2 x = 1\). This simplifies to \(\cos^2 x = \frac{1}{2}\), giving \(\cos x = \pm \frac{1}{\sqrt{2}}\). The solutions in the interval \(0 \leq x \leq \pi\) are \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).

(ii) (a) For \(k = 1\), \(f(x) = 3 - 4 \cos x\). The function is increasing from \((0, -1)\) to \((\pi, 7)\). It is not a straight line and flattens at the extremities, indicating an inflection.

(b) The function \(f\) has an inverse because it is one-to-one (1:1) or strictly increasing with no turning points.

(i) The function \(f(x) = 3 - 2 \tan\left(\frac{1}{2}x\right)\) is defined for \(0 \leq x < \pi\). The range of \(\tan\left(\frac{1}{2}x\right)\) is \((-\infty, \infty)\), so \(f(x)\) can take any value less than or equal to 3. Thus, the range is \(f \leq 3\).

(ii) To find \(f\left(\frac{2}{3}\pi\right)\), substitute \(x = \frac{2}{3}\pi\) into the function: \(f\left(\frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{2} \times \frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{3}\pi\right) = 3 - 2\sqrt{3}\).

(iii) The graph of \(y = f(x)\) starts at \(y = 3\) when \(x = 0\) and decreases as \(x\) increases, with no turning points, tending tangentially towards \(x = \pi\).

(iv) To find \(f^{-1}(x)\), start with \(y = 3 - 2 \tan\left(\frac{x}{2}\right)\). Rearrange to make \(x\) the subject: \(y = 3 - 2 \tan\left(\frac{x}{2}\right) \Rightarrow 2 \tan\left(\frac{x}{2}\right) = 3 - y \Rightarrow \tan\left(\frac{x}{2}\right) = \frac{3-y}{2} \Rightarrow \frac{x}{2} = \tan^{-1}\left(\frac{3-y}{2}\right) \Rightarrow x = 2 \tan^{-1}\left(\frac{3-y}{2}\right)\). Thus, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{3-x}{2}\right)\).

(i) Solve \(4 - 3 \sin x = 2\) to get \(3 \sin x = 2\), so \(\sin x = \frac{2}{3}\). The solutions are \(x = 0.730\) or \(x = 2.41\).

(ii) The graph of \(y = 4 - 3 \sin x\) should show one complete oscillation, correctly shaped and positioned in the first quadrant.

(iii) The equation \(f(x) = k\) has no solution for \(k < 1\) or \(k > 7\).

(iv) The largest value of \(A\) for which \(g\) has an inverse is \(A = \frac{3\pi}{2}\).

(v) For \(g^{-1}(3)\), solve \(3 = 4 - 3 \sin x\) to get \(\sin x = \frac{1}{3}\). Thus, \(g^{-1}(3) = 2.80\).

(i) The function \(f(x) = 5 - 3 \sin 2x\) is based on the sine function, which oscillates between -1 and 1. Therefore, \(-3 \leq -3 \sin 2x \leq 3\). Adding 5 to each part of the inequality gives \(2 \leq 5 - 3 \sin 2x \leq 8\). Thus, the range of \(f\) is \(2 \leq f(x) \leq 8\).

(ii) The graph of \(y = 5 - 3 \sin 2x\) will show one complete oscillation from \(x = 0\) to \(x = \pi\). The graph starts at \(y = 5\), decreases to \(y = 2\), and then increases back to \(y = 8\) at \(x = \pi\). It is important to note that the graph does not touch the x-axis and is in the first quadrant.

(iii) The function \(f\) does not have an inverse because it is not one-to-one. The function \(f(x) = 5 - 3 \sin 2x\) repeats values within the interval \(0 \leq x \leq \pi\), which means it fails the horizontal line test for invertibility.