Answer: \(\theta=15^\circ,165^\circ\); \(\phi=\frac{5\pi}{12},\frac{11\pi}{12},\frac{17\pi}{12},\frac{23\pi}{12}\).

Use

\(\operatorname{sec}\theta=\frac1{\cos\theta},\qquad \tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \operatorname{cosec}\theta=\frac1{\sin\theta}.\)

Then

\(\operatorname{sec}\theta-\frac{\tan\theta}{\operatorname{cosec}\theta} =\frac1{\cos\theta}-\frac{\frac{\sin\theta}{\cos\theta}}{\frac1{\sin\theta}}.\)

So

\(\frac1{\cos\theta}-\frac{\sin^2\theta}{\cos\theta} =\frac{1-\sin^2\theta}{\cos\theta} =\frac{\cos^2\theta}{\cos\theta} =\cos\theta.\)

Replacing \(\theta\) by \(2\theta\), the equation becomes

\(\cos2\theta=\frac{\sqrt3}{2}.\)

Since \(0^\circ\leq\theta\leq180^\circ\), we have \(0^\circ\leq2\theta\leq360^\circ\). Thus

\(2\theta=30^\circ\quad\text{or}\quad330^\circ.\)

Hence

\(\boxed{\theta=15^\circ,\ 165^\circ}.\)

For part (b),

\(2\sin^2\left(\phi+\frac{\pi}{3}\right)=1\)

gives

\(\sin^2\left(\phi+\frac{\pi}{3}\right)=\frac12.\)

Let \(u=\phi+\frac{\pi}{3}\). Since \(0\lt \phi\lt 2\pi\),

\(\frac{\pi}{3}\lt u\lt \frac{7\pi}{3}.\)

In this interval,

\(u=\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4}.\)

Subtract \(\frac{\pi}{3}\):

\(\phi=\boxed{\frac{5\pi}{12},\frac{11\pi}{12},\frac{17\pi}{12},\frac{23\pi}{12}}.\)