For b), just note that $11$ is a prime $4k+3$ so infinite descent finishes.

For part a) just take $a=11^x$ $b=11^y$ $c=11^z$ $d=11^t$ such that $2021x-2022z=1$ and $2023y-2024t=1$.

For part b),it's obvious that $11$ divides $a^{2022}+b^{2022}$,so we have two cases: 1. If $11 \mid a$ $\implies 11 \mid b$, then either $11 \mid c$,which will be a contradiction, or $11\nmid c$ and from Fermat itss easy to see this is not possible. 2.$11 \nmid a$ and this case can be disproved by using Fermat and looking for quadratic residues mod $11$.