When $x$is divided by 4 leaves the remainder 3, then the remainder when ${(2022 + x)}^{ 2022}$ is dividend by 8 is  

It is given that $x=4k+3,k\in N$

$\begin{array}{l}\therefore (2022+x{)}^{2022}\\ =\left\{\right(2020+2)+(4k+3){\}}^{2022}\\ =(2024+4k+1{)}^{2022}\\ =\left\{4\right(506+k)+1{\}}^{2020}\end{array}$

$=(4n+1{)}^{2020},$ where $n=506+k$

$=\left\{{ }^{2020}{C}_0(4n{)}^{2020}+\dots {+}^{2020}{C}_{2019}(4n)\right\}+1$

$=8\lambda +1,$ where $\lambda \in N$

Hence , $(2022+x{)}^{2022}$ when divided by 8 leaves the remainder 1.