(a) Start with the identity for tangent of a sum:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \tan 60^\circ}{1 - \tan \theta \tan 60^\circ}\)

Since \(\tan 60^\circ = \sqrt{3}\), substitute to get:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\)

Given \(\tan(\theta + 60^\circ) = 2 \cot \theta\), rewrite \(\cot \theta\) as \(\frac{1}{\tan \theta}\):

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} = \frac{2}{\tan \theta}\)

Cross-multiply to obtain:

\(\tan \theta (\tan \theta + \sqrt{3}) = 2(1 - \tan \theta \sqrt{3})\)

Expand and simplify:

\(\tan^2 \theta + \sqrt{3} \tan \theta = 2 - 2\sqrt{3} \tan \theta\)

Rearrange to form the quadratic equation:

\(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\)

(b) Solve the quadratic equation \(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\) using the quadratic formula:

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{(3\sqrt{3})^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\)

Calculate the discriminant:

\(\sqrt{27 + 8} = \sqrt{35}\)

Thus,

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{35}}{2}\)

Calculate the solutions for \(\theta\):

\(\tan \theta = 0.3599\) gives \(\theta = 19.8^\circ\)

\(\tan \theta = -5.556\) gives \(\theta = 100.2^\circ\)

Thus, the solutions are \(\theta = 19.8^\circ\) and \(\theta = 100.2^\circ\).