(i) Differentiate \(x = t^2 + 3t + 1\) to get \(\frac{dx}{dt} = 2t + 3\).

Differentiate \(y = t^4 + 1\) to get \(\frac{dy}{dt} = 4t^3\).

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t^3}{2t + 3}\).

Set \(\frac{dy}{dx} = 4\) and solve: \(\frac{4t^3}{2t + 3} = 4\).

Simplify to get \(4t^3 = 4(2t + 3)\).

\(t^3 = 2t + 3\).

Thus, \(p = \sqrt[3]{2p + 3}\).

(ii) Evaluate \(p - \sqrt[3]{2p + 3}\) at \(p = 1.8\) and \(p = 2.0\).

At \(p = 1.8\), \(1.8 - \sqrt[3]{2(1.8) + 3} \approx -0.076\).

At \(p = 2.0\), \(2.0 - \sqrt[3]{2(2.0) + 3} \approx 0.087\).

Since the sign changes, \(p\) is between 1.8 and 2.0.

(iii) Use the iterative formula \(p_{n+1} = \sqrt[3]{2p_n + 3}\).

Start with \(p_0 = 1.9\).

\(p_1 = \sqrt[3]{2(1.9) + 3} \approx 1.8854\).

\(p_2 = \sqrt[3]{2(1.8854) + 3} \approx 1.8895\).

\(p_3 = \sqrt[3]{2(1.8895) + 3} \approx 1.8899\).

\(p_4 = \sqrt[3]{2(1.8899) + 3} \approx 1.8899\).

The value of \(p\) correct to 2 decimal places is 1.89.