(i) Differentiate \(y = x^2 \cos \frac{1}{2}x\) using the product rule:

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

Set \(\frac{dy}{dx} = 0\) for a stationary point:

\(2x \cos \frac{1}{2}x = \frac{1}{2}x^2 \sin \frac{1}{2}x\).

Rearrange to get \(\tan \frac{1}{2}x = \frac{4}{x}\).

Thus, \(\tan \frac{1}{2}p = \frac{4}{p}\).

(ii) Calculate \(\tan \frac{1}{2}x\) at \(x = 2\) and \(x = 2.5\):

For \(x = 2\), \(\tan 1 = 1.5574\) and \(\frac{4}{2} = 2\).

For \(x = 2.5\), \(\tan 1.25 = 1.9207\) and \(\frac{4}{2.5} = 1.6\).

Since \(1.5574 < 2\) and \(1.9207 > 1.6\), \(p\) is between 2 and 2.5.

(iii) Use the iterative formula:

Start with \(p_0 = 2\).

\(p_1 = 2 \arctan \left( \frac{4}{2} \right) = 2 \arctan(2) = 2.0344\).

\(p_2 = 2 \arctan \left( \frac{4}{2.0344} \right) = 2.1489\).

\(p_3 = 2 \arctan \left( \frac{4}{2.1489} \right) = 2.1455\).

\(p_4 = 2 \arctan \left( \frac{4}{2.1455} \right) = 2.1456\).

\(p_5 = 2 \arctan \left( \frac{4}{2.1456} \right) = 2.1456\).

Thus, \(p \approx 2.15\) to 2 decimal places.