(i) Substitute \(x = u^2\), then \(dx = 2u \, du\). The integral becomes \(\int_0^{p^2} \cos(\sqrt{x}) \, dx = \int_0^p 2u \cos(u) \, du\).

Integrate by parts: let \(v = \cos(u)\) and \(dw = 2u \, du\), then \(dv = -\sin(u) \, du\) and \(w = u^2\).

\(\int 2u \cos(u) \, du = 2u \sin(u) + 2 \cos(u)\).

Using limits \(u = 0\) to \(u = p\), we have:

\(2p \sin(p) + 2 \cos(p) - (0 + 2) = 1\).

Simplifying gives \(2p \sin(p) = 3 - 2 \cos(p)\), hence \(\sin(p) = \frac{3 - 2 \cos(p)}{2p}\).

(ii) Using the iterative formula \(p_{n+1} = \sin^{-1} \left( \frac{3 - 2 \cos(p_n)}{2p_n} \right)\):

\(p_1 = 1.0000\)

\(p_2 = 1.2399\)

\(p_3 = 1.2497\)

\(p_4 = 1.2500\)

\(p_5 = 1.2500\)

Thus, \(p = 1.25\) correct to 2 decimal places.