(a) At \(\theta = 0\), \(r = a \cot\left(\frac{1}{3}\pi\right) = 2\sqrt{3}\). Solving gives \(a = 2\).

(b) The area is given by \(\frac{1}{2} \int_{0}^{\frac{1}{6}\pi} 4\cot^2\left(\frac{1}{3}\pi - \theta\right) \, d\theta\). This simplifies to \(2 \left[ \cot\left(\frac{1}{3}\pi - \theta\right) - \theta \right]_{0}^{\frac{1}{6}\pi}\), resulting in \(\frac{4}{3}\sqrt{3} - \frac{1}{3}\sqrt{3}\pi\).

(c) Using identities for \(\cos(A-B)\) and \(\sin(A-B)\), and substituting \(r = \sqrt{x^2 + y^2}\), \(x = r \cos \theta\), \(y = r \sin \theta\), the Cartesian equation is shown as \(2(x + y\sqrt{3}) = (x\sqrt{3} - y)\sqrt{x^2 + y^2}\).