(i) To solve \(fg(x) = 1\), substitute \(g(x) = \frac{1}{2}x\) into \(f(x)\):

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

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

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

\(x = 2 \times \cos^{-1}\left(\frac{1}{3}\right) \approx 2.46\)

Alternative: Solve \(f(y) = 1 \rightarrow y = 1.23 \rightarrow \frac{1}{2}x = 1.23 \rightarrow x = 2.46\)

(ii) The graph of \(y = f(x) = 2 - 3 \cos x\) is a cosine curve. Sketch one cycle from \(0\) to \(2\pi\), starting and ending at the same negative \(y\) value, with a peak at \(y = 5\) and a trough at \(y = -1\).