(i) Differentiate \(y = e^{-2x} \ln(x-1)\) using the product rule:

\(\frac{dy}{dx} = -2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1}\).

Set \(\frac{dy}{dx} = 0\):

\(-2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1} = 0\).

Simplify to find \(x = 1 + e^{\frac{1}{2(x-1)}}\).

(ii) Calculate \(f(x) = \ln(x-1) - \frac{1}{2(x-1)}\):

\(f(2.2) = -0.234\), \(f(2.6) = 0.317\).

Calculate \(f(x) = 2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1}\):

\(f(2.2) = 0.005\), \(f(2.6) = -0.0017\).

Since \(f(x)\) changes sign, \(p\) is between 2.2 and 2.6.

(iii) Use the iterative formula \(p_{n+1} = 1 + \exp\left( \frac{1}{2(p_n-1)} \right)\):

Start with \(p_0 = 2.4\).

\(p_1 = 2.4152\)

\(p_2 = 2.4213\)

\(p_3 = 2.4198\)

\(p_4 = 2.4204\)

\(p_5 = 2.4201\)

\(p_6 = 2.4203\)

\(p_7 = 2.4202\)

Thus, \(p \approx 2.42\) to 2 decimal places.