Answer: (a) \(\theta=15.9^\circ,\ -164.1^\circ,\ -45^\circ,\ 135^\circ\). (b) \(\phi=0.743,\ 1.30,\ 2.84\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

For part (a), factorise:

\(7\tan^2\theta+5\tan\theta-2=0\).

This gives

\((7\tan\theta-2)(\tan\theta+1)=0.\)

So \(\tan\theta=\frac27\) or \(\tan\theta=-1\).

In the interval \(-180^\circ\leqslant\theta\leqslant180^\circ\), this gives

\(\theta=15.9^\circ,\ -164.1^\circ,\ -45^\circ,\ 135^\circ.\)

For part (b),

\(3\sin(3\phi-1.5)-2=0\), so \(\sin(3\phi-1.5)=\frac23\).

Let \(u=3\phi-1.5\). Since \(0\lt\phi\lt3\), we have \(-1.5\lt u\lt7.5\).

The relevant solutions are

\(u=0.7297\ldots,\quad u=2.4119\ldots,\quad u=7.0129\ldots.\)

Therefore

\(\phi=\frac{u+1.5}{3}\), giving

\(\phi=0.743,\ 1.30,\ 2.84\) to 3 significant figures.

This gives the required answer: (a) \(\theta=15.9^\circ,\ -164.1^\circ,\ -45^\circ,\ 135^\circ\). (b) \(\phi=0.743,\ 1.30,\ 2.84\).