Answer: \(x=21.1^\circ,\ 81.1^\circ,\ 141.1^\circ\); \(y=0.430,\ \pi,\ 5.85\) radians.

For part (a)(i),

\((1-\sin A)(1+\sin A)=1-\sin^2A=\cos^2A.\)

Therefore

\(\dfrac{(1-\sin A)(1+\sin A)}{\sin A\cos A}=\dfrac{\cos^2A}{\sin A\cos A}=\dfrac{\cos A}{\sin A}=\operatorname{cot}A.\)

Using this identity with angle \(3x\), the equation becomes

\(\operatorname{cot}3x=\dfrac12.\)

So

\(\tan3x=2.\)

Since \(0^\circ\leq x\leq180^\circ\), we have \(0^\circ\leq3x\leq540^\circ\).

The solutions for \(3x\) in this interval are

\(3x=63.434^\circ,\ 243.434^\circ,\ 423.434^\circ.\)

Divide by \(3\):

\(x=21.145^\circ,\ 81.145^\circ,\ 141.145^\circ.\)

So

\(\boxed{x=21.1^\circ,\ 81.1^\circ,\ 141.1^\circ}.\)

For part (b), use \(\tan^2y=\operatorname{sec}^2y-1\):

\(10(\operatorname{sec}^2y-1)-\operatorname{sec}y-1=0.\)

So

\(10\operatorname{sec}^2y-\operatorname{sec}y-11=0.\)

Factorise:

\((10\operatorname{sec}y-11)(\operatorname{sec}y+1)=0.\)

Hence

\(\operatorname{sec}y=\dfrac{11}{10}\quad\text{or}\quad \operatorname{sec}y=-1.\)

So

\(\cos y=\dfrac{10}{11}\quad\text{or}\quad \cos y=-1.\)

For \(0\leq y\leq2\pi\), this gives

\(y=0.4297\ldots,\quad y=\pi,\quad y=2\pi-0.4297\ldots.\)

Therefore

\(\boxed{y=0.430,\ \pi,\ 5.85}\) radians.