(a) \[\frac{dx}{d\theta}=-6\sin2\theta,\qquad\frac{dy}{d\theta}=4\cos\theta.\] In the interior of the given interval, \[\frac{dy}{dx}=\frac{4\cos\theta}{-6\sin2\theta}=-\frac1{3\sin\theta}.\] At P the gradient is 2, so \(\sin\theta=-1/6\). Only the quadrant III value is in the interval: \[\theta=\pi+\sin^{-1}(1/6)\approx3.31\text{ radians}.\]

(b) Substituting the parametric equations into the line, \[9\cos2\theta+4\sin\theta=0.\] Let \(u=\sin\theta\), and use \(\cos2\theta=1-2u^2\): \[18u^2-4u-9=0,\qquad u=\frac{2\pm\sqrt{166}}{18}.\] Since \(-1\le\sin\theta\le0\) on the given interval, \[\sin\theta=\frac{2-\sqrt{166}}{18}\approx-0.604672.\] Hence the gradient at Q is \[\frac{dy}{dx}=-\frac1{3\sin\theta}=\frac6{\sqrt{166}-2}\approx0.551.\]