Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

a) 

If $9$ divides $a$, $9$ also divides its sum of digits. So, if $18$ divides a number, $9$ divides the sum of its digits. But, there are only $3$ digits whose sum is at most $27$. So, the sum of digits can be $9$, $18$ or $27$.

But, for $27$, the number has to be $999$. But, $18$ does not divide $999$. So, the only possible values are $9, 18$.

Example for $9$: $108$

Example for $18$: $198$

b)

In any consecutive $18$ numbers, one of them is divisible by $18$. Its sum is either $9$ or $18$ (from part a). $9$ and $18$ both divides $18$. So, its sum of digits divide itself.

So, we can find such a number and that is the number divisible by $18$.