(a) Calculate \(\cot \frac{1}{2}x - 3x\) at \(x = 0.5\) and \(x = 1\).

For \(x = 0.5\), \(\cot \frac{1}{2} \times 0.5 = \cot 0.25 \approx 3.92\) and \(3 \times 0.5 = 1.5\). So, \(3.92 - 1.5 > 0\).

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.83\) and \(3 \times 1 = 3\). So, \(1.83 - 3 < 0\).

Since there is a change of sign, \(\alpha\) lies between 0.5 and 1.

(b) Rearrange \(\cot \frac{x}{2} = 3x\) to \(x = \frac{1}{3} \left( x + 4 \arctan \left( \frac{1}{3x} \right) \right)\).

The iterative formula \(x_{n+1} = \frac{1}{3} \left( x_n + 4 \arctan \left( \frac{1}{3x_n} \right) \right)\) is derived from this rearrangement.

If the sequence converges, it must converge to the root \(\alpha\).

(c) Start with an initial guess, say \(x_0 = 0.5\).

Calculate successive iterations:

\(x_1 = \frac{1}{3} \left( 0.5 + 4 \arctan \left( \frac{1}{3 \times 0.5} \right) \right) \approx 0.7623\)

\(x_2 = \frac{1}{3} \left( 0.7623 + 4 \arctan \left( \frac{1}{3 \times 0.7623} \right) \right) \approx 0.8037\)

\(x_3 = \frac{1}{3} \left( 0.8037 + 4 \arctan \left( \frac{1}{3 \times 0.8037} \right) \right) \approx 0.7921\)

\(x_4 = \frac{1}{3} \left( 0.7921 + 4 \arctan \left( \frac{1}{3 \times 0.7921} \right) \right) \approx 0.7951\)

\(x_5 = \frac{1}{3} \left( 0.7951 + 4 \arctan \left( \frac{1}{3 \times 0.7951} \right) \right) \approx 0.7943\)

\(x_6 = \frac{1}{3} \left( 0.7943 + 4 \arctan \left( \frac{1}{3 \times 0.7943} \right) \right) \approx 0.7945\)

The value converges to \(0.79\) to 2 decimal places.