Answer: \(y=30^\circ,150^\circ,199.5^\circ,340.5^\circ\); \(z=\dfrac{7\pi}{24},\dfrac{11\pi}{24}\).

For part (a), use a common denominator:

\(\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x} = \frac{\sin^2x+(1+\cos x)^2}{\sin x(1+\cos x)}.\)

Expand the numerator:

\(\sin^2x+(1+\cos x)^2 = \sin^2x+1+2\cos x+\cos^2x.\)

Since \(\sin^2x+\cos^2x=1\), this becomes

\(2+2\cos x=2(1+\cos x).\)

Therefore

\(\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x} = \frac{2(1+\cos x)}{\sin x(1+\cos x)} = \frac2{\sin x} = 2\operatorname{cosec}x.\)

For part (b)(i), use \(\operatorname{cosec}^2y=1+\operatorname{cot}^2y\), so \(\operatorname{cot}^2y=\operatorname{cosec}^2y-1\).

Then

\(\operatorname{cosec}^2y-1+\operatorname{cosec}y-5=0.\)

So

\(\operatorname{cosec}^2y+\operatorname{cosec}y-6=0.\)

Factorise:

\((\operatorname{cosec}y-2)(\operatorname{cosec}y+3)=0.\)

Thus

\(\operatorname{cosec}y=2\quad\text{or}\quad \operatorname{cosec}y=-3.\)

So

\(\sin y=\frac12\quad\text{or}\quad \sin y=-\frac13.\)

For \(0^\circ\le y\le360^\circ\), this gives

\(y=30^\circ,\ 150^\circ,\ 199.5^\circ,\ 340.5^\circ.\)

For part (b)(ii), let

\(\theta=2z+\frac{\pi}{4}.\)

Since \(0\le z\le\pi\),

\(\frac{\pi}{4}\le \theta\le \frac{9\pi}{4}.\)

In this interval,

\(\cos\theta=-\frac{\sqrt3}{2}\)

when

\(\theta=\frac{5\pi}{6}\quad\text{or}\quad \theta=\frac{7\pi}{6}.\)

Therefore

\(2z+\frac{\pi}{4}=\frac{5\pi}{6} \quad\text{or}\quad 2z+\frac{\pi}{4}=\frac{7\pi}{6}.\)

So

\(2z=\frac{7\pi}{12}\quad\text{or}\quad 2z=\frac{11\pi}{12}.\)

Hence

\(\boxed{z=\frac{7\pi}{24}\ \text{or}\ z=\frac{11\pi}{24}}.\)