Answer: (a) \(33\); (b)(ii) \((2x-1)(3x+5)(x-1)\); (b)(iii) \(\theta=30^\circ,90^\circ\).

We start with the main method. Use the factor theorem first, then factorise the remaining quadratic factor.

For part (a), by the remainder theorem the remainder on division by \(x-2\) is \(\mathrm p(2)\).

\(\mathrm p(2)=6(2)^3+(2)^2-12(2)+5=48+4-24+5=33.\)

For part (b)(i), evaluate \(\mathrm p\left(\frac12\right)\):

\(\mathrm p\left(\frac12\right)=6\left(\frac12\right)^3+\left(\frac12\right)^2-12\left(\frac12\right)+5.\)

This is

\(\frac34+\frac14-6+5=0.\)

Therefore \(2x-1\) is a factor.

Dividing \(6x^3+x^2-12x+5\) by \(2x-1\) gives

\(3x^2+2x-5.\)

Now factorise:

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

Hence

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

For part (b)(iii), let \(u=\sin\theta\). Then

\(6u^3+u^2-12u+5=0\)

has factors

\((2u-1)(3u+5)(u-1)=0.\)

So

\(u=\frac12,\quad u=-\frac53,\quad u=1.\)

Since \(0^\circ\leqslant\theta\leqslant90^\circ\), \(u=-\frac53\) is impossible. Therefore

\(\sin\theta=\frac12\) gives \(\theta=30^\circ\), and \(\sin\theta=1\) gives \(\theta=90^\circ\).

This completes the solution and gives the required result.