(i) The area of the sector CMB is \(\frac{1}{2} a^2 \theta\).

The area of the triangle ABC is \(\frac{1}{2} a^2\).

Given that the area of the sector is one third of the area of the triangle:

\(\frac{1}{2} a^2 \theta = \frac{1}{3} \times \frac{1}{2} a^2\)

\(a^2 \theta = \frac{1}{3} a^2\)

\(\theta = \frac{1}{3}\)

Using the formula for the tangent of angle \(\theta\):

\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{a} = 1\)

Thus, \(\tan \theta = 3\theta\).

(ii) Using the iterative formula \(\theta_{n+1} = \arctan(3\theta_n)\):

Start with an initial guess, say \(\theta_0 = 1\).

\(\theta_1 = \arctan(3 \times 1) = \arctan(3) \approx 1.2490\)

\(\theta_2 = \arctan(3 \times 1.2490) = \arctan(3.7470) \approx 1.3084\)

\(\theta_3 = \arctan(3 \times 1.3084) = \arctan(3.9252) \approx 1.3235\)

\(\theta_4 = \arctan(3 \times 1.3235) = \arctan(3.9705) \approx 1.3259\)

\(\theta_5 = \arctan(3 \times 1.3259) = \arctan(3.9777) \approx 1.3265\)

Thus, the root correct to 2 decimal places is \(\theta = 1.32\).