Do you mean find max m?

The cases with $m\ge11$ are clearly impossible, as then two of $x_1,\ldots,x_m$ must leave the same remainder after division by 11, and hence ontradict (a). The case with $m=10$ is possible, as for instance the ten numbers $1,2,3,4,5,6,7,8,9,10$ satisfy both conditions.

nmd27082001 wrote: Do you mean find max m? yes

For $m=11$ just all $x$-es must be different reminders mod 11, thus one is divisible by $11$ (WLOG $x_1\equiv 0 \pmod{11}$), thus $x_2x_3...x_{11} \equiv 0\pmod{11}$ must hold: that is $1\cdot 2 \cdot ...\cdot 10 =10!\equiv 0 \pmod{11}$, which is not true by Wilson's theorem.