Answer: \(a=29,\ b=-9\); \(x=-3,\frac15\); \(\theta=99.7^\circ,170.3^\circ,279.7^\circ,350.3^\circ\).

We start with the main method. Apply the factor theorem: if \((x-a)\) is a factor, then \(p(a)=0\).

(a) Since \(x+3\) is a factor of both \(\mathrm p(x)\) and \(\mathrm p'(x)\),

\(\mathrm p(-3)=0\quad\text{and}\quad \mathrm p'(-3)=0.\)

Now

\(\mathrm p'(x)=15x^2+2ax+39.\)

Using \(\mathrm p'(-3)=0\):

\(135-6a+39=0.\)

So

\(174-6a=0,\)

hence

\(a=29.\)

Using \(\mathrm p(-3)=0\):

\(5(-3)^3+29(-3)^2+39(-3)+b=0.\)

This gives

\(-135+261-117+b=0,\)

so

\(9+b=0.\)

Hence

\(b=-9.\)

(b) With these values,

\(\mathrm p(x)=5x^3+29x^2+39x-9.\)

Since \(x+3\) is a factor, divide by \(x+3\):

\(\mathrm p(x)=(x+3)(5x^2+14x-3).\)

Then

\(5x^2+14x-3=(5x-1)(x+3).\)

Thus

\(\mathrm p(x)=(x+3)^2(5x-1).\)

So the solutions of \(\mathrm p(x)=0\) are

\(x=-3\quad\text{or}\quad x=\frac15.\)

(c) Let

\(u=\operatorname{cosec}2\theta.\)

The equation becomes

\(\mathrm p(u)=0,\)

so

\(u=-3\quad\text{or}\quad u=\frac15.\)

But \(\left|\operatorname{cosec}2\theta\right|\geq1\), so \(u=\frac15\) is impossible.

Therefore

\(\operatorname{cosec}2\theta=-3,\)

which gives

\(\sin2\theta=-\frac13.\)

For \(0^\circ\leq\theta\leq360^\circ\), we have \(0^\circ\leq2\theta\leq720^\circ\).

The solutions for \(2\theta\) are approximately

\(199.47^\circ,\quad340.53^\circ,\quad559.47^\circ,\quad700.53^\circ.\)

Hence

\(\theta=99.7^\circ,\quad170.3^\circ,\quad279.7^\circ,\quad350.3^\circ.\)

This completes the solution and gives the required result.