(i) The function \(f(x) = 2 - 3 \cos x\) has a range determined by the range of \(\cos x\), which is \([-1, 1]\). Therefore, the range of \(f(x)\) is \(2 - 3(-1) \leq f(x) \leq 2 - 3(1)\), which simplifies to \(-1 \leq f(x) \leq 5\).

(ii) The graph of \(y = f(x)\) is a cosine wave starting at \((0, -1)\), peaking at \((\pi, 5)\), and ending at \((2\pi, -1)\). It is symmetrical about \(x = \pi\).

(iii) For \(g(x) = 2 - 3 \cos x\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\cos x\) is one-to-one is \([0, \pi]\). Thus, the largest value of \(p\) is \(\pi\).

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

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

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

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

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