(a) The first and fifth terms sum to twice the third: \[2\cos x+10\cos x=-12\sqrt3\sin x.\] Thus \(\tan x=-1/\sqrt3\). In the given second quadrant, \[\boxed{x=5\pi/6}.\]

(b) The first term is \(a=2\cos(5\pi/6)=-\sqrt3\). The difference is \[d=\frac{10\cos x-2\cos x}{4}=2\cos x=-\sqrt3.\] Therefore \[\boxed{S_{25}=\frac{25}{2}[2(-\sqrt3)+24(-\sqrt3)]=-325\sqrt3}.\]