(a) Simplify \(\dfrac{(3+2\sqrt5)(6-2\sqrt5)}{4-\sqrt5}\), giving your answer in the form \(a+b\sqrt5\), where \(a\) and \(b\) are integers.

(b) Triangle \(ABC\) is such that \(AB=6-2\sqrt3\), \(BC=6+2\sqrt3\), and \(\cos ABC=-\dfrac12\). Find \(AC\), giving your answer in the form \(c\sqrt d\), where \(c\) and \(d\) are integers.