Remainders of large powers can be found by looking for the point where the remainders start to…

(a) $7^2 = 49$ and $7^4 = 2401 \equiv 1 \pmod{100}$. Since $2026 = 4 \times 506 + 2$, we get $7^{2026} \equiv 7^2 = 49 \pmod{100}$.

(b) Powers of $9$ modulo $100$: $9^2 = 81$, $9^4 = 6561 \equiv 61$, $9^5 \equiv 61 \times 9 = 549 \equiv 49$, $9^{10} \equiv 49^2 = 2401 \equiv 1$. So the cycle has length $10$. Now $9^9 \equiv (-1)^9 \equiv 9 \pmod{10}$, so $9^{9^9} \equiv 9^9 \pmod{100}$, and $9^9 = 9^5 \cdot 9^4 \equiv 49 \times 61 = 2989 \equiv 89 \pmod{100}$.

(c) Powers of $7$ modulo $1000$ return to $1$ first at $n = 20$, and no earlier. Any such $n$ must divide $100$, the cycle length for modulus $1000$, and $7^{20} \equiv 1 \pmod{1000}$ by direct multiplication.

(d) Modulo $100$ the powers of $7$ return to $1$ exactly when $4 \mid n$. There are $\lfloor 2026 / 4 \rfloor = 506$ such $n$.

(e) $2026 = 12 \times 168 + 10$, so $2^{2026} \equiv 2^{10} = 1024 \equiv 10 \pmod{13}$. Also $3^3 \equiv 1$ and $2026 = 3 \times 675 + 1$, so $3^{2026} \equiv 3$. The sum is $10 + 3 = 0 \pmod{13}$.