Oxidization and reduction chemistry questions

1. Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction:

a) HgO(s) --> Hg(l) + O2(g); Hg is reduced and O is oxidized

b) Al(s) + MnO2(s) --> Al2O3(s) + Mn(s); Al is oxidized and Mn is reduced

-These two I attempted to do but idk if they are correct.

2.Use the half reaction method to write balanced ionic equations for each reaction. All are basic solutions!

a) MnO4-(aq) + ClO2-(aq) --> MnO2(s) + ClO4-(aq)

b) Mn3+(aq) + I-(aq) --> Mn2+(aq) + IO3-(aq)

3. Identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent in each unbalanced redox equation. Then balance each redox equation by the oxidation-number-change method.

a) Cu(s) + HNO3(aq) --> Cu(NO3)2(aq) + NO2(g) + H2O(l)

b) Bi(OH)3(s) + Na2SnO2(aq) --> Bi(s) + Na2SnO3(aq) + H2O(l)

c) P(s) + HNO3(aq) + H2O(l) --> NO(g) + H3PO4(aq)

These I have no idea what to do.

oxidation occurs when an element loses electrons.....Loes Electrons Oxidized
reduction occurs wehn an element gaines electrons.....Gain Electron Reduced
LEO the lion goes GER

HgO(s) --> Hg(l) + O2(g)
Hg in HgO = 2+ and becomes Hg = 0, reduced, gained electrons
O in HgO = 2-, becomes neutral in O2, oxidized, lose electrons

Al(s) + MnO2(s) --> Al2O3(s) + Mn(s)
Al is neutral as a solid and becomes 3+ in Al2O3, oxidized, lost electrons
Mn in MnO2 = 4+, becomes 0 as Mn(s), reduced, gained electrons

2. balance in base
MnO4- + ClO2- --> MnO2 + ClO4-
Mn in MnO4- = 7+ becomes 4+ in MnO2.....reduced
Cl in ClO2- = 3+ becomes 7+ in ClO4-.....oxidized

MnO4- + 3e- + 4H2O --> MnO2 + 2H2O + 4OH-
ClO2- + 2H2O + 4OH- --> ClO4- + 4e- + 4H2O
multiply through by common factor to make electrons the same
4MnO4- + 12 e- + 12H2O --> 4MnO2 + 8H2O + 16OH-
3ClO2- + 6H2O + 12OH- --> 3ClO4- + 12e- + 12H2O