(a) The curve is sketched with \(r\) strictly increasing from \(\theta = 0\) to \(\theta = 2\). The maximum distance is found by evaluating \(r\) at \(\theta = 2\), giving \(r = \sqrt{\frac{1}{2}\arctan(1)} = \frac{1}{2}\sqrt{\pi}\).

(b) The area is calculated using the integral \(\frac{1}{2} \int_0^2 \arctan\left(\frac{1}{2}\theta\right) d\theta\). Integration by parts is applied:

\(\frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \int \frac{\theta}{\theta^2 + 4} d\theta \right]_0^2\)

\(= \frac{1}{2}\left[ \frac{1}{2}\theta\arctan\left(\frac{1}{2}\theta\right) - \ln(\theta^2 + 4) \right]_0^2\)

\(= \frac{1}{4}\pi - \frac{1}{2}\ln 2\)

(c) To find the point furthest from \(\theta = \frac{1}{2}\pi\), set \(x = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \cos\theta\) and differentiate:

\(\frac{dx}{d\theta} = (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} \sin\theta + \cos\theta (\arctan\left(\frac{1}{2}\theta\right))^{\frac{1}{4}} (\theta^2 + 4)^{-1} = 0\)

\((\theta^2 + 4)\arctan\left(\frac{1}{2}\theta\right)\sin\theta - \cos\theta = 0\)

Check sign change between 0.6 and 0.7:

\((0.6^2 + 4)\arctan(0.3)\sin 0.6 - \cos 0.6 = -0.108\)

\((0.7^2 + 4)\arctan(0.35)\sin 0.7 - \cos 0.7 = 0.209\)

Thus, a root exists between 0.6 and 0.7.