(i) The function is given by \(f(x) = a + b \cos 2x\). We have two conditions:

\(f(0) = a + b \cos 0 = -1\) which simplifies to \(a + b = -1\).

\(f\left(\frac{1}{2}\pi\right) = a + b \cos \pi = 7\) which simplifies to \(a - b = 7\).

Solving these two equations:

\(a + b = -1\)

\(a - b = 7\)

Add the equations: \(2a = 6\) so \(a = 3\).

Substitute \(a = 3\) into \(a + b = -1\):

\(3 + b = -1\) so \(b = -4\).

Thus, \(a = 3\) and \(b = -4\).

(ii) To find the x-coordinates where the curve intersects the x-axis, set \(f(x) = 0\):

\(3 - 4 \cos 2x = 0\)

\(\cos 2x = \frac{3}{4}\)

Solving for \(x\), we find:

\(2x = \cos^{-1}\left(\frac{3}{4}\right)\)

\(x = \frac{1}{2} \cos^{-1}\left(\frac{3}{4}\right)\)

Calculating gives \(x \approx 0.36\) and \(x \approx 2.78\).

(iii) The graph of \(y = f(x)\) is a cosine wave starting at \(y = -1\) when \(x = 0\) and reaching a maximum of \(y = 7\) at \(x = \frac{1}{2}\pi\), then returning to \(y = -1\) at \(x = \pi\). The graph oscillates once between \(x = 0\) and \(x = \pi\).