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 $18$.