Okay, so the half reactions start out as:
MnO4 + 3e- ----> MnO2
NO2- -----> NO3- + 2e-
The next thing you need to do is balance the Oxygens
MnO4 + 3e- ----> MnO2 + 2H2O
H2O + NO2- -----> NO3- + 2e-
Now you need to balance the Hs by adding H plusses to the other side (we will get rid of these later with the base)
4H+ + MnO4 + 3e- ----> MnO2 + 2H2O
H2O + NO2- -----> NO3- + 2e- + 2H+
Since the Electrons don't balance, we need to multiply to get a common denominator
2(4H+ + MnO4 + 3e- ----> MnO2 + 2H2O)
3(H2O + NO2- -----> NO3- + 2e- + 2H+)
Now we can put the equations together and cancel the electrons out
8H+ + 2MnO4 + 6e- + 3H2O + 3NO2- -----> 2MnO2 + 4H2O + 3NO3- + 6e- + 6 H+
Now we need to cancel out the H+ with -OH. We should add 8 because that will cancel out all of the -OH on both sides (2 are left over on the right side). You need to make sure that you add it to both sides :) When the 2 combine, it makes H2O
8H2O + 2MnO4 + 3H2O +3NO2- ----> 2MnO2 + 4H2O + 3NO3- + 6H2O + 2 OH-
Now the last thing you need to do is cancel out the extra H2Os and you're done!
H2O + 2MnO4 + 3NO2- ----> 2MnO2 + 4H2O + 3NO3- + 2 OH-
Okay, so the half reactions start out as:
MnO4 + 3e- ----> MnO2
NO2- -----> NO3- + 2e-
Wrong - you don't know number of electrons at this stage. You use electrons to balance charge AFTER oxygen and hydrogen is balanced. Also you have missed some charges (MnO4-) in the equations you have wrote.
See half reaction method (http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method) for detailed method explanation.
H2O + 2MnO4 + 3NO2- ----> 2MnO2 + 4H2O + 3NO3- + 2 OH-
That's hardly balanced - you have water on both sides. And canceling it won't help, it will be still not balanced.