Answer: (a) \(x=11.5^\circ\) or \(x=168.5^\circ\).

(a) Use the identity

\(\operatorname{cot}^2x=\operatorname{cosec}^2x-1.\)

Then

\(3(\operatorname{cosec}^2x-1)-14\operatorname{cosec}x-2=0.\)

Simplify:

\(3\operatorname{cosec}^2x-14\operatorname{cosec}x-5=0.\)

Let \(u=\operatorname{cosec}x\). Then

\(3u^2-14u-5=0.\)

Factorise:

\((3u+1)(u-5)=0.\)

So

\(u=-\frac13\quad\text{or}\quad u=5.\)

The value \(u=-\frac13\) is impossible for \(\operatorname{cosec}x\), since \(|\operatorname{cosec}x|\geq1\). Therefore

\(\operatorname{cosec}x=5,\)

so

\(\sin x=\frac15.\)

For \(0^\circ\lt x\lt 360^\circ\), sine is positive in quadrants I and II. Hence

\(x=11.5^\circ\quad\text{or}\quad x=168.5^\circ.\)

Therefore

\(\boxed{x=11.5^\circ\text{ or }168.5^\circ}.\)

(b) Start with the left hand side:

\(\frac{\sin^4y-\cos^4y}{\operatorname{cot} y}.\)

Using the difference of two squares,

\(\sin^4y-\cos^4y=(\sin^2y-\cos^2y)(\sin^2y+\cos^2y)=\sin^2y-\cos^2y.\)

Also \(\operatorname{cot} y=\frac{\cos y}{\sin y}\), so

\(\frac{\sin^4y-\cos^4y}{\operatorname{cot} y}=(\sin^2y-\cos^2y)\frac{\sin y}{\cos y}.\)

Hence

\(=\frac{\sin^3y}{\cos y}-\sin y\cos y.\)

Now

\(\tan y-2\cos y\sin y=\frac{\sin y}{\cos y}-2\sin y\cos y.\)

Putting this over a common denominator gives

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

Since \(1-2\cos^2y=\sin^2y-\cos^2y\), this is

\((\sin^2y-\cos^2y)\frac{\sin y}{\cos y}.\)

This is the same as the left hand side, so

\(\boxed{\frac{\sin^4y-\cos^4y}{\operatorname{cot} y}=\tan y-2\cos y\sin y}.\)