(a) This is the same as computing 4⁵⁵ (mod 11). We have
4² ≡ 16 ≡ 5 (mod 11)
4³ ≡ 4 • 5 ≡ 20 ≡ 9 (mod 11)
4⁴ ≡ 4 • 9 ≡ 36 ≡ 3 (mod 11)
4⁵ ≡ 4 • 3 ≡ 12 ≡ 1 (mod 11)
Then from here,
4⁵⁵ ≡ (4⁵)¹⁰ • 4⁵ ≡ 1¹⁰ • 1⁵ ≡ 1 (mod 11)
(b) Each term in the sum
4ⁿ + 4ⁿ⁺¹ + 4ⁿ⁺² + 4ⁿ⁺³ + 4ⁿ⁺⁴
has a common factor of 4ⁿ, so this sum is the same as
4ⁿ (1 + 4 + 16 + 64 + 256) = 4ⁿ • 341 = 4ⁿ • 11 • 31
So the sum is indeed divisible by 11 for all integers n.
(c) Since 4ⁿ = (2ⁿ)², we know the sum is also divisible by 11 when a = 2.
