(i) Consider the function \(f(x) = x - \frac{10}{e^{2x} - 1}\).

Calculate \(f(1) = 1 - \frac{10}{e^2 - 1}\) and \(f(2) = 2 - \frac{10}{e^4 - 1}\).

Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, \(\alpha\) lies between 1 and 2.

(ii) Assume the sequence \(x_n\) converges to \(L\). Then \(L = \frac{1}{2} \ln \left( 1 + \frac{10}{L} \right)\).

Rearranging gives \(L = \frac{10}{e^{2L} - 1}\), so \(L = \alpha\).

(iii) Use the iterative formula:

Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{1}{2} \ln \left( 1 + \frac{10}{1} \right) = 1.1513\).

Calculate \(x_3 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1513} \right) = 1.1394\).

Calculate \(x_4 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1394} \right) = 1.1427\).

Calculate \(x_5 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1427} \right) = 1.1414\).

Calculate \(x_6 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1414} \right) = 1.1418\).

Since the values are converging, \(\alpha \approx 1.14\) to 2 decimal places.