a) no, because numbers give different remainders b) $c=7,d=0$ then $7022...231=79*88...89$

Letting the size of the block $\overline{22\cdots 231}$ to be $n$, it boils down checking whether for $p\in\{73,79\}$, \[ p\mid 200\frac{10^n-1}{9} + 31 + \overline{xy}\cdot 10^n \implies p\mid 10^n(9\cdot \overline{xy}+2)+79 \]where $\overline{xy} \in\{\overline{ab},\overline{cd}\}$. From here, we readily obtain (b) as then $p\mid 9X+2$ is indeed satisfied for $c=7$ and $d=0$. As for (a), note that if $9\overline{xy}+2\equiv 0\pmod{73}$ then we have a contradiction. If not then for this fixed $a,b$, $10^n\equiv 10^{n+1}\pmod{73}$, which is clearly absurd.