(a) The curve is a spiral shape on \([0, 2\pi]\), starting at the pole with one other point of intersection with the initial line. The maximum distance from the pole is \(a\).

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

\(A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta = \frac{1}{2} a^2 \int_0^{2\pi} \tan^2\left(\frac{1}{8}\theta\right) \, d\theta\)

Using \(\tan^2 A = \sec^2 A - 1\), we have:

\(= \frac{1}{2} a^2 \left[ 8 \tan\left(\frac{1}{8}\theta\right) - \theta \right]_0^{2\pi}\)

\(= \frac{1}{2} a^2 (8 - 2\pi) = a^2 (4 - \pi)\)

(c) Let \(y = a \tan\left(\frac{1}{8}\theta\right) \sin\theta\). The derivative is:

\(\frac{dy}{d\theta} = a \left( \tan\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta \sec^2\left(\frac{1}{8}\theta\right) \right) = 0\)

\(\Rightarrow \cos\left(\frac{1}{8}\theta\right) \sin\left(\frac{1}{8}\theta\right) \cos\theta + \frac{1}{8} \sin\theta = 0\)

\(\Rightarrow \sin\left(\frac{1}{4}\theta\right) \cos\theta + \frac{1}{4} \sin\theta = 0\)

\(\Rightarrow 4 \sin\left(\frac{1}{4}\theta\right) \cos\theta + \sin\theta = 0\)

Checking values:

\(4 \sin\left(\frac{1}{4} \times 4.95\right) \cos 4.95 + \sin 4.95 \approx -0.0822\)

\(4 \sin\left(\frac{1}{4} \times 5\right) \cos 5 + \sin 5 \approx 0.118\)

There is a sign change, confirming a root between 4.95 and 5.