Redox equations in Chemistry

So I am completely lost when it comes to this oxidation process. I understand that H is +1, O is -2 and Monoatomic ions = the ion charge and elemental atoms are 0. But when the equations get really long and complex I get lost. How do I identify the oxidation number for those without rules?
More precisely these equations:

(a) HCl(aq) + HNO3(aq) + Sn(s) → NO2(g) + H2SnCl6(aq) + H2O(l)
-oxidizing agent & reducing agent

(b) PbO(s) + NH3(aq) → N2(g) + H2O(l) + Pb(s)
-oxidizing & reducing agents

(c) HCl(aq) + HNO3(aq) + Cu(s) → NO(g) + CuCl2(aq) + H2O(l)
-oxidizing & reducing agents

(d) I2(s) + Na2S2O3(aq) → Na2S4O6(aq) + NaI(aq)
-oxidizing & reducing agents

Your explanation as to how you found the oxidation numbers is greatly appreciated! Thanks for the help!!

The sum of the charges in a compound = 0
When Cl, or any other element in column 17, is the only negative ion in a compound, the charge = -1
Elements is column 1 have +1 charge
Elements in column 2 have +2 charge
H is +1, unless it is in a metallic hydride.
In sodium hydride, NaH, Na+ 1, H = -1

a) HCl(aq) + HNO3(aq) + Sn(s) → NO2(g) + H2SnCl6(aq) + H2O(l)

-oxidizing agent & reducing agent

H = +1, so Cl = -1
H = +1, 3 O’s = -6, so N = +5
Sn = 0

2 O’s = -4, so N = +4
2 H’s = +2, 6 Cl = -6, so Sn = +4
H = +1 and O = -2

The N+5 is reduced to N+4, as the Sn 0 is oxidized to Sn+4
Sn 0 is the reducing agent and HNO3 is the oxidizing agent.

(b) PbO(s) + NH3(aq) → N2(g) + H2O(l) + Pb(s)
-oxidizing & reducing agents
Pb+2 + O -2 = 0
H+3 + N -3 = 0
N2 = 0
Pb = 0
Pb +2 is reduced to Pb 0, as N-3 is oxidized to N 0
PbO is the oxidizing agent and NH3 is the reducing agent.

(c) HCl(aq) + HNO3(aq) + Cu(s) → NO(g) + CuCl2(aq) + H2O(l)
-oxidizing & reducing agents
N+5 is reduced to N+2
Cu 0 is oxidized to Cu+2