(i) Integrate \(x \cos x\) using integration by parts:

Let \(u = x\) and \(dv = \cos x \, dx\). Then \(du = dx\) and \(v = \sin x\).

\(\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C\).

Evaluate from 0 to \(a\):

\([x \sin x + \cos x]_0^a = a \sin a + \cos a - (0 \cdot \sin 0 + \cos 0) = a \sin a + \cos a - 1\).

Set equal to 0.5:

\(a \sin a + \cos a - 1 = 0.5\)

\(a \sin a = 1.5 - \cos a\)

\(\sin a = \frac{1.5 - \cos a}{a}\)

(ii) Verify \(a > 1\):

Consider \(a = 1\):

\(1 \sin 1 + \cos 1 - 1 \approx 0.8415 + 0.5403 - 1 = 0.3818\)

\(0.3818 < 0.5\), so \(a > 1\).

(iii) Iterative formula:

Start with \(a_1 = 1.2\).

\(a_2 = \sin^{-1} \left( \frac{1.5 - \cos 1.2}{1.2} \right) \approx 1.2464\)

\(a_3 = \sin^{-1} \left( \frac{1.5 - \cos 1.2464}{1.2464} \right) \approx 1.2461\)

\(a_4 = \sin^{-1} \left( \frac{1.5 - \cos 1.2461}{1.2461} \right) \approx 1.2461\)

Final answer: \(a \approx 1.2461\)