(i) Differentiate \(y = x^2 \cos 2x\) using the product rule: \(y' = 2x \cos 2x - 2x^2 \sin 2x\). Set \(y' = 0\) to find the maximum point. Solving \(2x \cos 2x - 2x^2 \sin 2x = 0\) gives \(\cos 2x = x \sin 2x\). Rearranging, \(\tan 2x = \frac{1}{x}\). At \(x = p\), \(2p = \arctan \left( \frac{1}{p} \right)\), so \(p = \frac{1}{2} \arctan \left( \frac{1}{p} \right)\).

(ii) Use the iterative formula \(p_{n+1} = \frac{1}{2} \arctan \left( \frac{1}{p_n} \right)\). Start with \(p_0 = 0.5\):

\(p_1 = 0.55357\)

\(p_2 = 0.53261\)

\(p_3 = 0.54070\)

\(p_4 = 0.53755\)

Final answer: \(p = 0.54\) to 2 decimal places.

(iii) Integrate \(y = x^2 \cos 2x\) by parts. Let \(u = x^2\) and \(dv = \cos 2x \, dx\), then \(du = 2x \, dx\) and \(v = \frac{1}{2} \sin 2x\). The integration by parts gives:

\(\int x^2 \cos 2x \, dx = \frac{1}{2} x^2 \sin 2x - \int x \sin 2x \, dx\).

Integrate \(\int x \sin 2x \, dx\) by parts again. Let \(u = x\) and \(dv = \sin 2x \, dx\), then \(du = dx\) and \(v = -\frac{1}{2} \cos 2x\). This gives:

\(\int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \, dx\).

Complete the integration to obtain:

\(\frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x\).

Substitute limits \(x = 0\) and \(x = \frac{1}{4} \pi\):

\(\frac{1}{32}(\pi^2 - 8)\).