(a) Expand and use the double-angle identities: \[\begin{aligned}4\cos\theta\sin(\theta+30^\circ)&=2\sqrt3\cos\theta\sin\theta+2\cos^2\theta\\&=\sqrt3\sin2\theta+\cos2\theta+1.\end{aligned}\] Matching \(R\cos(2\theta-\alpha)=R\cos\alpha\cos2\theta+R\sin\alpha\sin2\theta\), we obtain \(R\cos\alpha=1\), \(R\sin\alpha=\sqrt3\). Thus \[R=2,\qquad\alpha=60^\circ,\qquad k=1,\] and the required form is \(2\cos(2\theta-60^\circ)+1\).

(b) Apply part (a) with \(\theta=2\phi\), noting the coefficient is multiplied by 3: \[3\bigl(2\cos(4\phi-60^\circ)+1\bigr)=5\Rightarrow\cos(4\phi-60^\circ)=\frac13.\] Since \(-60^\circ<4\phi-60^\circ<300^\circ\), the relevant cosine angles are \(70.5288\ldots^\circ\) and \(289.4712\ldots^\circ\). Therefore \[\phi=32.6^\circ\quad\text{or}\quad87.4^\circ.\]