(a) Sketch the graphs of \(y = \cot \frac{1}{2}x\) and \(y = 1 + e^{-x}\). The intersection of these graphs in the interval \(0 < x \leq \pi\) shows there is exactly one root.

(b) Calculate the values at \(x = 1\) and \(x = 1.5\):

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.8305\) and \(1 + e^{-1} \approx 1.3679\).

For \(x = 1.5\), \(\cot \frac{1}{2} \times 1.5 = \cot 0.75 \approx 1.3764\) and \(1 + e^{-1.5} \approx 1.2231\).

The change in sign between these values confirms a root between 1 and 1.5.

(c) Using the iterative formula:

Start with an initial guess, say \(x_0 = 1\).

\(x_1 = 2 \arctan \left( \frac{1}{1 + e^{-1}} \right) \approx 1.3524\)

\(x_2 = 2 \arctan \left( \frac{1}{1 + e^{-1.3524}} \right) \approx 1.3402\)

\(x_3 = 2 \arctan \left( \frac{1}{1 + e^{-1.3402}} \right) \approx 1.3398\)

\(x_4 = 2 \arctan \left( \frac{1}{1 + e^{-1.3398}} \right) \approx 1.3398\)

The root correct to 2 decimal places is 1.34.