(i) The function \(f(x) = 4 \sin x - 1\) has a maximum value when \(\sin x = 1\) and a minimum value when \(\sin x = -1\). Therefore, the range of \(f(x)\) is:

\(4(-1) - 1 \leq f(x) \leq 4(1) - 1\)

\(-5 \leq f(x) \leq 3\)

(ii) To find the x-intercept, set \(f(x) = 0\):

\(4 \sin x - 1 = 0\)

\(\sin x = \frac{1}{4}\)

\(x = \sin^{-1}\left(\frac{1}{4}\right) \approx 0.253\)

So, the x-intercept is \((0.253, 0)\).

For the y-intercept, set \(x = 0\):

\(f(0) = 4 \sin 0 - 1 = -1\)

So, the y-intercept is \((0, -1)\).

(iii) The graph of \(y = f(x)\) is a sinusoidal curve starting at \((0, -1)\), peaking at \((\frac{1}{2}\pi, 3)\), and troughing at \((-\frac{1}{2}\pi, -5)\).

(iv) To find the inverse, solve \(y = 4 \sin x - 1\) for \(x\):

\(y + 1 = 4 \sin x\)

\(\sin x = \frac{y + 1}{4}\)

\(x = \sin^{-1}\left(\frac{y + 1}{4}\right)\)

Thus, \(f^{-1}(x) = \sin^{-1}\left(\frac{x + 1}{4}\right)\).

The domain of \(f^{-1}\) is the range of \(f\), \(-5 \leq x \leq 3\), and the range of \(f^{-1}\) is the domain of \(f\), \(-\frac{1}{2}\pi \leq f^{-1}(x) \leq \frac{1}{2}\pi\).

(i) Start with the equation \(5 + 3 \cos\left(\frac{1}{2}x\right) = 7\). Subtract 5 from both sides to get \(3 \cos\left(\frac{1}{2}x\right) = 2\). Divide by 3: \(\cos\left(\frac{1}{2}x\right) = \frac{2}{3}\). Find \(\frac{1}{2}x = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.84\). Multiply by 2 to solve for \(x\): \(x \approx 1.68\).

(ii) The graph of \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\) is a cosine wave starting at \(y = 8\) when \(x = 0\) and ending at \(y = 2\) when \(x = 2\pi\).

(iii) The function \(f\) has no turning points and is one-to-one over the given interval, so it has an inverse.

(iv) To find \(f^{-1}(x)\), start with \(y = 5 + 3 \cos\left(\frac{1}{2}x\right)\). Rearrange to make \(x\) the subject: \(y - 5 = 3 \cos\left(\frac{1}{2}x\right)\), then \(\cos\left(\frac{1}{2}x\right) = \frac{y-5}{3}\). Solve for \(x\): \(\frac{1}{2}x = \cos^{-1}\left(\frac{y-5}{3}\right)\), so \(x = 2 \cos^{-1}\left(\frac{y-5}{3}\right)\).

(i) Start with the equation \(6 - 4\cos\left(\frac{1}{2}x\right) = 4\). Simplify to \(4\cos\left(\frac{1}{2}x\right) = 2\), giving \(\cos\left(\frac{1}{2}x\right) = \frac{1}{2}\). The solutions for \(\frac{1}{2}x = \frac{1}{3}\pi\) and \(\frac{1}{2}x = \frac{5}{3}\pi\) lead to \(x = \frac{2}{3}\pi\) and \(x = \frac{10}{3}\pi\). However, \(x = \frac{10}{3}\pi\) is outside the given domain, so the valid solutions are \(x = \frac{2}{3}\pi\) and \(x = \frac{4}{3}\pi\).

(ii) The range of \(f(x) = 6 - 4\cos\left(\frac{1}{2}x\right)\) is determined by the range of \(\cos\left(\frac{1}{2}x\right)\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(6 - 4(1) = 2\) to \(6 - 4(-1) = 10\), so \(2 \leq f(x) \leq 10\).

(iii) The graph of \(y = f(x)\) is a cosine wave with a point of inflexion at \(x = \pi\).

(iv) To find \(f^{-1}(x)\), start with \(y = 6 - 4\cos\left(\frac{1}{2}x\right)\). Rearrange to \(\cos\left(\frac{1}{2}x\right) = \frac{6-y}{4}\). Then, \(\frac{1}{2}x = \cos^{-1}\left(\frac{6-y}{4}\right)\), leading to \(x = 2\cos^{-1}\left(\frac{6-y}{4}\right)\). Therefore, \(f^{-1}(x) = 2\cos^{-1}\left(\frac{6-x}{4}\right)\).

(i) To solve \(f(x) = 0\), we have:

\(3 \cos x - 2 = 0\)

\(\cos x = \frac{2}{3}\)

Using the inverse cosine function, \(x = \cos^{-1} \left( \frac{2}{3} \right)\), which gives \(x \approx 0.841\) or \(x \approx 5.44\) within the interval \(0 \leq x \leq 2\pi\).

(ii) The range of \(f(x) = 3 \cos x - 2\) is determined by the range of \(\cos x\), which is \([-1, 1]\). Therefore, the range of \(f(x)\) is:

\(3(-1) - 2 \leq f(x) \leq 3(1) - 2\)

\(-5 \leq f(x) \leq 1\)

(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = 1\), decreasing to \(y = -5\), and returning to \(y = 1\) over one cycle from \(0\) to \(2\pi\).

(iv) For \(g\) to have an inverse, it must be one-to-one. The maximum interval for \(x\) where \(\cos x\) is one-to-one is \([0, \pi]\). Therefore, \(k = \pi\) or \(180^\circ\).

(v) To find \(g^{-1}(x)\), solve \(y = 3 \cos x - 2\) for \(x\):

\(y + 2 = 3 \cos x\)

\(\cos x = \frac{y + 2}{3}\)

\(x = \cos^{-1} \left( \frac{y + 2}{3} \right)\)

Thus, \(g^{-1}(x) = \cos^{-1} \left( \frac{x + 2}{3} \right)\).