(a) To find the point furthest from the pole, differentiate \(r\) with respect to \(\theta\):

\(\frac{dr}{d\theta} = 2(\cos \theta) - (1 + \sin \theta) \sin \theta = 0\)

Simplifying gives:

\(1 - s - 2s^2 = 0\)

Solving the quadratic in \(\sin \theta\), we find \(\sin \theta = -1\) or \(\sin \theta = \frac{1}{2}\), leading to \(\theta = \frac{1}{6} \pi\).

The polar coordinates are \(\left( \frac{3}{2} \sqrt{3}, \frac{1}{6} \pi \right)\).

(b) The sketch of C shows a correct shape, tangential to \(\theta = \frac{\pi}{2}\) at the pole, with \(r\) strictly increasing to \(\frac{\pi}{6}\) then decreasing. The curve is in the first quadrant.

(c) The area of the region bounded by C and the initial line is found using:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} 4c^2(1+s)^2 \, d\theta\)

Using double angle formulae and integrating, we obtain:

\(\left[ \frac{5}{4} \theta + \frac{1}{2} \sin 2\theta - \frac{4}{3} \cos^3 \theta - \frac{1}{16} \sin 4\theta \right]_{0}^{\frac{1}{2}\pi}\)

The exact area is \(\frac{5}{8} \pi + \frac{4}{3}\).