(a) There are three identical Es: \[\frac{8!}{3!}=\boxed{6720}.\]

(b) Treat LED as a single fixed-order block. Together with R, A, S and two identical Es, this makes six objects: \[\frac{6!}{2!}=\boxed{360}.\]

(c) If A and D are adjacent, treat them as a block with two internal orders. The seven resulting objects include three identical Es, giving \(2(7!)/3!=1680\) arrangements. Therefore \[P(\text{A and D not together})=1-\frac{1680}{6720}=\boxed{\frac34}.\]