How to write net ionic equations for basic and acidic solutions?

Zn ------> Zn+2 & 2 electrons lost

(NO3)- & 8 electrons taken ------> (NH4)+
(N+5 in nitrate goes to N-3 in ammonium)

four times the Zn reaction to balance the electrons taken & lost:
4Zn ------> 4Zn+2 & 8 e-
(NO3)- & 8 e- ------> (NH4)+

combine the two 1/2 reactions:
4 Zn & (NO3)-1 ---> 4 Zn+2 & (NH4)+1

we need to balance the charges on both sides: -1 +9,....add 10 (OH-)'s to the right side & you will have 1- -1 :
4 Zn & (NO3)-1 ---> 4 Zn+2 & (NH4)+1 & 10 OH-

balance the H's using neutral water:
4 Zn & (NO3)-1 & 7 H2O ---> 4 Zn+2 & (NH4)+1 & 10 OH-

Also, here is a set-by-set way of doing them: http://tiny.cc/i0i6c

you can get answers or check answers to balanced redox reactions by going to this site!
http://www.gregthatcher.org//Chemistry/BalanceChemicalEquations.aspx?StartsWith=O

Mg (s) + VO43- (aq) Mg2+ (aq) + V2+ (aq) (Base)
Mg Mg2+ + 2 e-                 x3
3 e- + 8 H+ + VO43- V2+ + 4 H2O              x2
3Mg 3Mg2+ + 6 e-
6 e- + 16 H+ + 2 VO43- 2 V2+ + 8 H2O
3Mg + 16 H+ + 2 VO43- 3Mg2+ + 2 V2+ + 8 H2O + 16 OH-          on both sides

3Mg + 8 H2O + 2 VO43- 3Mg2+ + 2 V2+ + 16 OH