1) Cl2 + 2e¯ ---> 2Cl¯
2) Sn ---> Sn2+ + 2e¯
3) Fe2+ ---> Fe3+ + e¯
4) I3¯ + 2e¯ ---> 3I¯
5) ICl2¯ + 2e¯ ---> I¯ + 2Cl¯
6) Sn ---> SnO2 and NO3¯ ---> NO2
7) HClO ---> Cl2 and Co ---> Co2+
8) NO2 ---> NO3¯ and NO2 ---> NO
Note that in this last redox redox equation the NO2 (actually just the N) is both being reduced AND being oxidized. In the first half-reaction, the N goes from +4 to +5 (oxidation) and in the second, the N goes from +4 to +2 (reduction).
By the way, we could flip the reaction so that NO3¯ and NO are reacting together to produce only one product, the NO2. In that case, the two half-reactions would be reversed.