(i) To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt} = \frac{2}{t+2}\) and \(\frac{dy}{dt} = 3t^2 + 2\).

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 + 2}{\frac{2}{t+2}} = \frac{1}{2}(3t^2 + 2)(t+2)\).

At the origin, \(x = 0\) and \(y = 0\), which gives \(t = -1\).

Substitute \(t = -1\) into \(\frac{dy}{dx}\) to get \(\frac{5}{2}\) as the gradient at the origin.

(ii)(a) Given \(\frac{dy}{dx} = \frac{1}{2}\), equate to \(\frac{1}{2}(3p^2 + 2)(p+2)\) and solve to confirm \(p = \frac{1}{3p^2 + 2} - 2\).

(ii)(b) Use the iterative formula based on \(p = \frac{1}{3p^2 + 2} - 2\) to find \(p\).

Start with an initial guess, iterate to find \(p = -1.924\) or better \(-1.92367\).

Calculate coordinates using \(x = 2 \ln(p + 2)\) and \(y = p^3 + 2p + 3\).

Coordinates of \(P\) are \((-5.15, -7.97)\).