(i) To find the stationary point, we first find the derivative of \(y = \ln x + \frac{2}{x}\). The derivative is \(\frac{dy}{dx} = \frac{1}{x} - \frac{2}{x^2}\).

Setting \(\frac{dy}{dx} = 0\), we have \(\frac{1}{x} - \frac{2}{x^2} = 0\).

Multiplying through by \(x^2\), we get \(x - 2 = 0\), so \(x = 2\).

Substituting \(x = 2\) back into the original equation, \(y = \ln 2 + \frac{2}{2} = \ln 2 + 1\).

To determine if it is a minimum or maximum, consider the second derivative: \(\frac{d^2y}{dx^2} = -\frac{1}{x^2} + \frac{4}{x^3}\).

At \(x = 2\), \(\frac{d^2y}{dx^2} = -\frac{1}{4} + \frac{4}{8} = \frac{1}{4} > 0\), indicating a minimum point.

(ii) The iterative formula is \(x_{n+1} = \frac{2}{3 - \ln x_n}\). As \(n \to \infty\), \(x_n \to \alpha\), so \(\alpha = \frac{2}{3 - \ln \alpha}\).

Rearranging gives \(3 = \ln \alpha + \frac{2}{\alpha}\), which matches the equation \(y = 3\).

(iii) Using the iterative formula:

\(x_1 = 1\)

\(x_2 = \frac{2}{3 - \ln 1} = \frac{2}{3} \approx 0.67\)

\(x_3 = \frac{2}{3 - \ln 0.67} \approx 0.58\)

\(x_4 = \frac{2}{3 - \ln 0.58} \approx 0.56\)

\(x_5 = \frac{2}{3 - \ln 0.56} \approx 0.56\)

Thus, \(\alpha \approx 0.56\) to 2 decimal places.