A couple of things first: the MN locus is co-dominant, meaning that both can be expressed (first link).
The ABO blood group is the same way, except that O means the absence of either A or B (second link).
The Rh group is a single locus, and the Rh- allele is recessive (third link)
So, knowing all this, the mother has to be (AA or AO)/MM/(Rh+Rh- or Rh+Rh+) to get her own phenotype. However, we already know that the child is O, so the mother can't be AA and thus has to be AO; that is, AO/MM/(Rh+Rh- or Rh+Rh+). Again, since the child is Rh- the mother would have to be Rh+Rh-, so she is:
AO/MM/Rh+Rh-
If the child is O, MN, Rh- then given the dominance patterns we know that any father has to be (AO or OO)/(MN or NN)/(Rh+Rh- or Rh-Rh-). So the first father is AO/NN/Rh+Rh- and the second one is OO/MN/Rh-Rh-
So, doing Punnett squares, the first father is 25% likely to give an O (OO) child, while the second father is 50% likely.
Similarly, the N (NN) father is 100% likely to give an MN child; the MN father is 50% likely.
Finally, the Rh+Rh- father will give any Rh- children 25% of the time and the second father will have odds of 50%.
So, overall:
1st father = 25% * 100% * 25% = 12.5%
2nd father = 50% * 50% * 50% = 12.5%
Check my math and genotypes, but I think I did this right.