(a) REQUIREMENT contains two Rs, three Es and six other distinct letters: \[\frac{11!}{2!3!}=\boxed{3326400}.\]

(b) Treat RR and EEE as blocks. Together with the six other letters, there are eight distinct objects: \[8!=\boxed{40320}.\]

(c) The Rs occupy positions four apart. There are seven possible pairs, from \((1,5)\) to \((7,11)\). The remaining nine letters include three identical Es: \[7\frac{9!}{3!}=\boxed{423360}.\]

(d) Identical copies of a letter do not create new selections. The possibilities are EER with two other distinct letters, EERR with one other letter, EEER with one other letter, and EEERR. Thus \[\binom62+\binom61+\binom61+1=15+6+6+1=\boxed{28}.\]