Answer: \(x=0^\circ,38.0^\circ,90^\circ,128^\circ,180^\circ\), and \(y=-4.59,-0.947,1.69\).

(a) Factorise the equation:

\(\tan^2(2x)-4\tan(2x)=\tan(2x)(\tan(2x)-4)=0\).

So

\(\tan(2x)=0\quad\text{or}\quad \tan(2x)=4\).

Since \(0^\circ\leqslant x\leqslant180^\circ\), we have

\(0^\circ\leqslant2x\leqslant360^\circ\).

For \(\tan(2x)=0\),

\(2x=0^\circ,180^\circ,360^\circ\),

so

\(x=0^\circ,90^\circ,180^\circ\).

For \(\tan(2x)=4\),

\(2x=\tan^{-1}4=75.963\ldots^\circ\)

or, adding \(180^\circ\),

\(2x=255.963\ldots^\circ\).

Thus

\(x=37.981\ldots^\circ\quad\text{or}\quad x=127.981\ldots^\circ\).

To 3 significant figures, the solutions are

\(x=0^\circ,38.0^\circ,90^\circ,128^\circ,180^\circ\).

(b) Since

\(\operatorname{cosec}(y+1.2)=4\),

we have

\(\sin(y+1.2)=\frac14\).

Let

\(u=y+1.2\).

The given interval \(-5\lt y\lt2\) becomes

\(-3.8\lt u\lt3.2\).

The principal value is

\(\sin^{-1}\left(\frac14\right)=0.25268\ldots\).

In the interval \(-3.8\lt u\lt3.2\), the solutions are

\(u=0.25268\ldots,\quad u=\pi-0.25268\ldots,\quad u=-\pi-0.25268\ldots\).

Subtracting \(1.2\) from each value gives

\(y=-0.947\ldots,\quad y=1.688\ldots,\quad y=-4.594\ldots\).

Therefore, to 3 significant figures,

\(y=-4.59,-0.947,1.69\).