Answer: \(\theta=233.1^\circ\) approximately; \(\phi=0.132,\ 1.18\) approximately.

Using \(\operatorname{cosec}\theta=\frac1{\sin\theta}\) and \(\operatorname{cot}\theta=\frac{\cos\theta}{\sin\theta}\),

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

So

\(\frac{\operatorname{cosec}\theta-\operatorname{cot}\theta}{\sin\theta} =\frac{1-\cos\theta}{\sin^2\theta}.\)

Since \(\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)\),

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

Hence the equation becomes

\(\frac1{1+\cos\theta}=\frac52.\)

Therefore

\(2=5(1+\cos\theta),\)

so

\(\cos\theta=-\frac35.\)

In the interval \(180^\circ\lt \theta\lt 360^\circ\), this gives the third-quadrant solution

\(\boxed{\theta=233.1^\circ}\)

approximately.

For part (b), let

\(u=3\phi-4.\)

Then

\(\tan u=-\frac12.\)

Since \(0\lt \phi\lt \frac{\pi}{2}\),

\(-4\lt u\lt \frac{3\pi}{2}-4.\)

The relevant values of \(u\) in this interval are

\(u=-3.605\ldots,\quad -0.4636\ldots.\)

Hence

\(\phi=\frac{u+4}{3}.\)

This gives

\(\boxed{\phi=0.132,\ 1.18}\)

to 3 significant figures.