(i) Differentiate \(y = x \ln(8 - x)\) using the product rule: \(\frac{dy}{dx} = \ln(8 - x) - \frac{x}{8 - x}\).

Set \(\frac{dy}{dx} = 1\) and solve: \(\ln(8 - x) - \frac{x}{8 - x} = 1\).

Rearrange to find \(x = 8 - \frac{8}{\ln(8 - x)}\).

(ii) Calculate \(8 - \frac{8}{\ln(8 - 2.9)} = 3.09\) and \(8 - \frac{8}{\ln(8 - 3.1)} = 2.97\).

Since \(3.09 > 2.9\) and \(2.97 < 3.1\), \(a\) lies between 2.9 and 3.1.

(iii) Use the iterative formula \(x_{n+1} = 8 - \frac{8}{\ln(8 - x_n)}\).

Starting with \(x_0 = 3\), iterate: \(x_1 = 3.0293\), \(x_2 = 3.0111\), \(x_3 = 3.0223\), \(x_4 = 3.0154\), \(x_5 = 3.0198\).

Consecutive values \(3.0223\) and \(3.0154\) round to 3.02, confirming \(a = 3.02\).

(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.

(i) To differentiate \(y = \csc x = \frac{1}{\sin x}\), use the quotient rule or chain rule. The derivative of \(\sin x\) is \(\cos x\), so:

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{\sin x} \right) = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x.\)

(ii) The gradients of the curves are equal when \(x = a\). For \(y = \csc x\), the gradient is \(-\csc a \cot a\). For \(y = e^{-x}\), the gradient is \(-e^{-a}\). Equating these gives:

\(-e^{-a} = -\csc a \cot a.\)

Rearranging gives:

\(a = \arctan \left( \frac{e^a}{\sin a} \right).\)

(iii) To verify \(a\) lies between 1 and 1.5, calculate the expression for \(a = 1\) and \(a = 1.5\). Check that the values satisfy the inequality.

(iv) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{e^{a_n}}{\sin a_n} \right)\). Start with an initial guess, such as \(a_0 = 1\), and iterate:

\(a_1 = \arctan \left( \frac{e^{1}}{\sin 1} \right) \approx 1.317\)

Continue iterating until the value stabilizes to 3 decimal places:

\(a_2 \approx 1.317\)

\(a_3 \approx 1.317\)

The final answer is \(a \approx 1.317\).

(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.

(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 \((-5.15, -7.97)\).