For the second one, I think it's just a redox rxn.
so you got MnO4- -> MnO2, so if you balance that half rxn, it would be
3e- + 4H+ + MnO4- -> MnO2 + 2H2O
you do same for the I- -> IO3-,
3H2O + I- -> IO3- + 6H+ + 6e-.
Then you multiple the MnO4 half rxn coefficients by 2, to match up the 6 e's from the other half rxn.
Then add them up, to get
2H+ + I- + 2MnO4- -> IO3- + 2MnO2 + H2O.
May have done something wrong as I was doing it on the computer w/o writing them out, but the idea should be good 🙂