(a) The curve is sketched as a semicircle with \(\theta = 0\) at the rightmost point. The polar coordinates of the point furthest from the pole are \((1, 0)\).

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_{0}^{\pi} \frac{1}{\theta^2 + 1} \, d\theta\)

Integrating, we have:

\(\frac{1}{2} \left[ \arctan \theta \right]_{0}^{\pi} = \frac{1}{2} \arctan \pi = 0.631\)

(c) Let \(y = \frac{\sin \theta}{\sqrt{\theta^2 + 1}}\). The derivative is:

\(\frac{dy}{d\theta} = \left( \theta^2 + 1 \right)^{-\frac{1}{2}} \cos \theta - \theta \left( \theta^2 + 1 \right)^{-\frac{3}{2}} \sin \theta = 0\)

Solving gives:

\(\theta \neq 0 \Rightarrow \left( \theta + \frac{1}{\theta} \right) \cot \theta - 1 = 0\)

Checking for a root between 1.1 and 1.2:

\(\left( 1.1 + \frac{1}{1.1} \right) \cot 1.1 - 1 = 0.02 \ldots\)

\(\left( 1.2 + \frac{1}{1.2} \right) \cot 1.2 - 1 = -0.209 \ldots\)

There is a sign change, confirming a root exists between 1.1 and 1.2.