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