(i) Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{2}{3} \left( 1 + \frac{1}{1^2} \right) = \frac{2}{3} \times 2 = \frac{4}{3} \approx 1.3333\).

Calculate \(x_3 = \frac{2}{3} \left( 1.3333 + \frac{1}{(1.3333)^2} \right) \approx 1.2639\).

Calculate \(x_4 = \frac{2}{3} \left( 1.2639 + \frac{1}{(1.2639)^2} \right) \approx 1.2600\).

Calculate \(x_5 = \frac{2}{3} \left( 1.2600 + \frac{1}{(1.2600)^2} \right) \approx 1.2599\).

Since the values are converging, \(\alpha \approx 1.26\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\).

Solving \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\) gives \(\alpha^3 = 2\).

Thus, \(\alpha = \sqrt[3]{2}\) or \(2^{1/3}\).