Answer: (a)(ii) \(x=-75^\circ,-15^\circ,15^\circ,75^\circ\); (b) \(y=\frac{\pi}{2},\frac{11\pi}{6}\).

(a)(i) Start with

\(\frac{1}{\operatorname{sec}\theta-1}-\frac{1}{\operatorname{sec}\theta+1}.\)

Use a common denominator:

\(\frac{(\operatorname{sec}\theta+1)-(\operatorname{sec}\theta-1)}{(\operatorname{sec}\theta-1)(\operatorname{sec}\theta+1)}.\)

The numerator is \(2\), and the denominator is

\(\operatorname{sec}^2\theta-1.\)

Using \(\operatorname{sec}^2\theta-1=\tan^2\theta\),

\(\frac{1}{\operatorname{sec}\theta-1}-\frac{1}{\operatorname{sec}\theta+1} =\frac{2}{\tan^2\theta}.\)

Since \(\operatorname{cot}\theta=\frac1{\tan\theta}\),

\(\frac{2}{\tan^2\theta}=2\operatorname{cot}^2\theta.\)

(a)(ii) Using part (a)(i) with \(\theta=2x\),

\(\frac{1}{\sec2x-1}-\frac{1}{\sec2x+1}=2\operatorname{cot}^2 2x.\)

So

\(2\operatorname{cot}^2 2x=6.\)

Hence

\(\operatorname{cot}^2 2x=3,\)

so

\(\tan^2 2x=\frac13.\)

Therefore

\(\tan2x=\pm\frac1{\sqrt3}.\)

Since \(-90^\circ\lt x\lt90^\circ\),

\(-180^\circ\lt2x\lt180^\circ.\)

Thus

\(2x=-150^\circ,-30^\circ,30^\circ,150^\circ.\)

So

\(x=-75^\circ,-15^\circ,15^\circ,75^\circ.\)

(b) Since

\(\operatorname{cosec}\left(y+\frac{\pi}{3}\right)=2,\)

we have

\(\sin\left(y+\frac{\pi}{3}\right)=\frac12.\)

Let

\(u=y+\frac{\pi}{3}.\)

As \(0\leq y\leq2\pi\),

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

In this interval, \(\sin u=\frac12\) gives

\(u=\frac{5\pi}{6}\quad\text{or}\quad u=\frac{13\pi}{6}.\)

Therefore

\(y=\frac{5\pi}{6}-\frac{\pi}{3}=\frac{\pi}{2},\)

or

\(y=\frac{13\pi}{6}-\frac{\pi}{3}=\frac{11\pi}{6}.\)