Answer
See answer below.
Work Step by Step
a) Na: +1 in $NaI$ and in $Na_2SO_4$
I: -1 in $NaI$ and 0 at $I_2$
H: +1 in $H_2O$ and in $H_2SO_4$
S: +6 in $Na_2SO_4$, $H_2SO_4$ and $MnSO_4$
O:-2 in $H_2O$, $MnO_2$, $Na_2SO_4$, $H_2SO_4$ and $MnSO_4$
Mn: +4 in $MnO_2$ and +2 in $MnSO_4$
b) $NaI$ was oxidized since iodine's oxidation number increases, thus it's the reducing agent.
$MnO_2$ was reduced since manganese's oxidation number decreases, thus it's the oxidation agent.
c) From the picture we conclude it's product favored.
d) From left to right:
Sodium iodide, sulfuric acid, manganese (IV) oxide, sodium sulfate, manganese (II) sulfate, iodine gas, water.
