9 is the central number if the set size is odd; it's the average of the two central numbers if the set size is even. The mode is greater than the median so we need at least two 12s in the set.
If the set size is 5 then we have a, b, 9, 12, 12. The sum of all the data=5×9=45, so a+b=45-33=12.
a and b must be less than 9 and cannot be equal: 4+8, 5+7 are candidates. The set could be:
5, 7, 9, 12, 12.
There are many other possibilities.