(i) Sketch the graphs of \(y = \ln(x+2)\) and \(y = 4e^{-x}\). The graph of \(y = \ln(x+2)\) is a logarithmic curve starting from \(x = -2\) and increasing. The graph of \(y = 4e^{-x}\) is an exponential decay curve starting from \(y = 4\) when \(x = 0\) and decreasing. These graphs intersect at exactly one point, indicating one real root.

(ii) Calculate \(\ln(1+2) = \ln(3) \approx 1.0986\) and \(4e^{-1} \approx 1.4715\). Since \(\ln(3) < 4e^{-1}\), the function \(\ln(x+2) - 4e^{-x}\) changes sign between \(x = 1\) and \(x = 1.5\), confirming a root in this interval.

(iii) Using the iterative formula:

1. Start with \(x_0 = 1\).

2. \(x_1 = \ln\left( \frac{4}{\ln(1 + 2)} \right) \approx \ln\left( \frac{4}{1.0986} \right) \approx 1.2809\).

3. \(x_2 = \ln\left( \frac{4}{\ln(1.2809 + 2)} \right) \approx \ln\left( \frac{4}{1.2809} \right) \approx 1.2345\).

4. \(x_3 = \ln\left( \frac{4}{\ln(1.2345 + 2)} \right) \approx \ln\left( \frac{4}{1.2345} \right) \approx 1.2301\).

The root correct to 2 decimal places is 1.23.