How to take the beginning and end parts of a list with simpler syntax?

Update: You can also try MapAt:

a = Range[10];
a = MapAt[b &, a, {{;; 3}, {7 ;;}}]

{b, b, b, 4, 5, 6, b, b, b, b}

Or ReplaceAll

a = Range[10];
a /. Alternatives @@ Drop[a, 4;;6] -> b

{b, b, b, 4, 5, 6, b, b, b, b}

Original answer: Try Drop:

Drop[Range @ 10, {4, 6}]

{1, 2, 3, 7, 8, 9, 10}

Drop[Range @ 10, 4 ;; 6]

{1, 2, 3, 7, 8, 9, 10}

Drop[CharacterRange["a", "j"], {4, 6}]

{"a", "b", "c", "g", "h", "i", "j"}

If it is okay to perform two assignments instead of one, then we can write:

(a[[#]] = b) & /@ {1 ;; 3, 7 ;; 10}

Scan is probably better since it does not bother constructing the result list that we are just going to discard anyway:

Scan[(a[[#]] = b) &, {1 ;; 3, 7 ;; 10}]

Perhaps this?:

ReplacePart[Range@10, i_ /; Not[4 <= i <= 6] :> b]
(*  {b, b, b, 4, 5, 6, b, b, b, b}  *)