(a) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the shaded segment is \(\frac{1}{2} r^2 (\theta - \sin \theta)\). Given that the shaded area is one third of the semicircle's area, we have:

\(\frac{1}{2} r^2 (\theta - \sin \theta) = \frac{1}{3} \times \frac{1}{2} \pi r^2\)

\(\theta - \sin \theta = \frac{1}{3} \pi\)

Using \(\theta = \pi - 2\theta\), we substitute:

\(\theta = \frac{1}{3}(\pi - 1.5 \sin 2\theta)\)

(b) Evaluate the expression at \(\theta = 0.5\) and \(\theta = 0.7\):

For \(\theta = 0.5\): \(\frac{1}{3}(\pi - 1.5 \sin 1) \approx 0.626\)

For \(\theta = 0.7\): \(\frac{1}{3}(\pi - 1.5 \sin 1.4) \approx 0.554\)

Since \(0.5 < 0.626\) and \(0.7 > 0.554\), \(0.5 < \theta < 0.7\).

(c) Using the iterative formula \(\theta_{n+1} = \frac{1}{3}(\pi - 1.5 \sin 2\theta_n)\):

Start with \(\theta_0 = 0.6\):

\(\theta_1 = 0.58899\)

\(\theta_2 = 0.58641\)

\(\theta_3 = 0.58606\)

\(\theta_4 = 0.58602\)

\(\theta_5 = 0.58601\)

Thus, \(\theta \approx 0.586\) to 3 decimal places.