1. We need the remainder of $2021^{2022}+2022^{2021}$ when divided by $7$.

2. Reduce the bases modulo $7$: $2021 \equiv 2021-7\cdot 288 = 2021-2016 = 5 \pmod{7}$ $2022 \equiv 2022-7\cdot 288 = 2022-2016 = 6 \pmod{7}$ So, $2021^{2022}+2022^{2021} \equiv 5^{2022}+6^{2021} \pmod{7}.$

3. Now simplify each term.

For $5^{2022}$ modulo $7$: Since $5^2=25 \equiv 4 \pmod{7},$ $5^4 \equiv 4^2=16 \equiv 2 \pmod{7},$ $5^6 \equiv 5^4\cdot 5^2 \equiv 2\cdot 4=8 \equiv 1 \pmod{7}.$ Thus powers of $5$ repeat with period $6$ modulo $7$. Now, $2022 \equiv 0 \pmod{6},$ so $5^{2022} \equiv 5^0 \equiv 1 \pmod{7}.$

4. For $6^{2021}$ modulo $7$: $6 \equiv -1 \pmod{7}$ Therefore, $6^{2021} \equiv (-1)^{2021} = -1 \equiv 6 \pmod{7}.$

5. Add the two remainders: $5^{2022}+6^{2021} \equiv 1+6 = 7 \equiv 0 \pmod{7}.$

6. Therefore, the remainder is $\boxed{0}.$

7. Checking options:

• A: $0$ ✅
• B: $1$ ❌
• C: $2$ ❌
• D: $6$ ❌

Hence, the correct option is A.