Do not use a calculator in this question.

The triangle \(ABC\) has \(AB=4\sqrt3-5\), \(BC=4\sqrt3+5\) and \(\angle ABC=60^\circ\).

It is known that \(\sin60^\circ=\dfrac{\sqrt3}{2}\), \(\cos60^\circ=\dfrac12\), \(\tan60^\circ=\sqrt3\).

(i) Find the exact value of \(AC\).

(ii) Hence show that

\(\operatorname{cosec}ACB=\frac{2\sqrt p}{q}(4\sqrt3+5),\)

where \(p\) and \(q\) are integers.