(a) Start with the Cartesian equation \((x^2 + y^2)^2 = 36(x^2 - y^2)\).

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), so \(x^2 + y^2 = r^2\).

Substitute: \(r^4 = 36(r^2 \cos^2 \theta - r^2 \sin^2 \theta)\).

Factor out \(r^2\): \(r^4 = 36r^2(\cos^2 \theta - \sin^2 \theta)\).

Use the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\): \(r^2 = 36 \cos 2\theta\).

(b) The sketch is a closed curve symmetrical about the initial line. The maximum distance from the pole is 6.

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} 36 \cos 2\theta \, d\theta\).

\(= 18 \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos 2\theta \, d\theta\).

\(= 18 [\frac{1}{2} \sin 2\theta]_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\).

\(= 18 [\sin \frac{1}{2}\pi - \sin(-\frac{1}{2}\pi)]\).

\(= 18 [1 - (-1)] = 18 \times 2 = 18\).

(d) To find the maximum distance from the initial line, set \(y = 6 \cos^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve:

\(\cos^2 2\theta \cos \theta - \cos^{\frac{1}{2}} 2\theta \sin 2\theta \sin \theta = 0\).

Use trigonometric identities to solve for \(\theta\):

\(\theta = \pm \frac{1}{4}\pi\).

The maximum distance is \(3\sqrt{2}\).