Let the number ( 22 ) ^ 2022 + ( 2022 ) ^ 22 leave the remainder α...

Given that $\alpha$ be the remainder when ${\left(22\right)}^{2022} + {\left(2022\right)}^{22}$ is divided by $3$ and $\beta$ be the remainder when the same is divided by $7$.

$⇒ {\left(22\right)}^{2022} + {\left(2022\right)}^{22} = {\left(21 + 1\right)}^{2022} + {\left(2022\right)}^{22}$.

Here ${\left(2022\right)}^{22}$ is divisible by $3$ as $2022$ is divisible by $3$.

So on expanding ${\left(21 + 1\right)}^{2022}$, we get

$⇒ {\left(21 + 1\right)}^{2022} = C0 2022 {\left(21\right)}^{2022} + C1 2022 {\left(21\right)}^{2021} + . . . . . . + C2022 2022 {\left(1\right)}^{2022}$

$= 3\left({3}^{2021} \times {7}^{2022} + C1 2022 \times {3}^{2020} \times {7}^{2021} + . . . . . .\right) + 1$

$= 3{k}_{1} + 1$

In this case the remainder is $1$

Hence, $\alpha = 1$

Now, $⇒ {\left(22\right)}^{2022} + {\left(2022\right)}^{22} = {\left(21 + 1\right)}^{2022} + {\left(2023 - 1\right)}^{22}$

Take ${\left(2023 - 1\right)}^{22}$

$⇒ {\left(2023 - 1\right)}^{22} = C0 22 {\left(2023\right)}^{22} - C1 22 {\left(2023\right)}^{21} + . . . . . . + C22 22 {\left(- 1\right)}^{22}$

$= 7\left(C0 22 {\left(7\right)}^{21}{\left(289\right)}^{22} - C1 22 {\left(7\right)}^{20}{\left(289\right)}^{21} + . . . . . .\right) + 1$

$= 7{k}_{2} + 1$

$⇒ {\left(21 + 1\right)}^{2022} + {\left(2023 - 1\right)}^{22} = 7{k}_{1} + 1 + 7{k}_{2} + 1$

$= 7\mu + 2$

$⇒ \beta = 2$.

Hence, ${\alpha}^{2} + {\beta}^{2} = {1}^{2} + {2}^{2} = 5$.

Therefore, the required answer is $5$.