(i) Given the maximum value of \(f(x) = a - b \cos x\) is 10, and the minimum value is \(-2\), we have:

\(a + b = 10\) (when \(\cos x = -1\))

\(a - b = -2\) (when \(\cos x = 1\))

Solving these equations simultaneously:

\(a + b = 10\)

\(a - b = -2\)

Adding the equations gives:

\(2a = 8 \Rightarrow a = 4\)

Substituting \(a = 4\) into \(a + b = 10\):

\(4 + b = 10 \Rightarrow b = 6\)

(ii) To solve \(f(x) = 0\):

\(4 - 6 \cos x = 0\)

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

Using the inverse cosine function, \(x = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\) or \(360^\circ - 48.2^\circ = 311.8^\circ\).

(iii) The sketch of \(y = f(x)\) should show one cycle of a cosine wave, starting at \(-2\), peaking at 10, and returning to \(-2\) at \(x = 360^\circ\).