1. We are given that when $x$ is divided by $4$, the remainder is $3$.

   So, $x = 4k+3$ for some integer $k$.

2. We need the remainder when $(2020+x)^{2022}$ is divided by $8$.

3. First simplify the base modulo $8$.

   Since 2020 mod 8 = 4, we get $2020+x = 2020 + (4k+3) = 4k+2023.$

   Now modulo $8$: $2020+x \equiv 4 + x \pmod{8}.$

   But $x=4k+3$, so modulo $8$, $x$ can be either:

   • $3$ if $k$ is even,
   • $7$ if $k$ is odd.

   Hence: $2020+x \equiv 4+3=7 \pmod{8}$ or $2020+x \equiv 4+7=11 \equiv 3 \pmod{8}.$

   So the base is congruent to either $3$ or $7$ modulo $8$.

4. Now raise to the power $2022$.

   Since $2022$ is even, we use: $3^2=9 \equiv 1 \pmod{8},$ and $7^2=49 \equiv 1 \pmod{8}.$

   Therefore, any even power of $3$ or $7$ is congruent to $1$ modulo $8$.

   So, $3^{2022} \equiv 1 \pmod{8},$ $7^{2022} \equiv 1 \pmod{8}.$

5. Hence in both possible cases, $(2020+x)^{2022} \equiv 1 \pmod{8}.$

Therefore, the remainder is: $\boxed{1}$