Half reaction method redox reaction ?

1.
Oxidation half equation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Reduction half equation: NO₂⁻(aq) + 2H⁺(aq) + e⁻ → NO(g) + H₂O(ℓ)

(Oxidation half equation) + (Reduction half equation), and cancel e⁻ on the both sides. The overall redox equation is:
Fe²⁺(aq) + NO₂⁻(aq) + 2H⁺(aq) → Fe³⁺(aq) + H₂O(ℓ)

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2.
Oxidation half equation: SO₂(g) + 2H₂O(ℓ) → SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻
Reduction half equation: ClO₃⁻(aq) + 6H⁺(aq) + 6e⁻ → Cl⁻(aq) + 3H₂O(ℓ)

(Oxidation half equation)×3 + (Reduction half equation)×1, and cancel 6e⁻, 3H₂O(ℓ) and 6H⁺(aq) on the both sides. The overall redox equation is:
ClO₃⁻(aq) + 3SO₂(g) + 3H₂O(ℓ) → Cl⁻(aq) + 3SO₄²⁻(aq) + 6H⁺(aq)

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3.
Oxidation half equation: NO₂⁻(aq) + H₂O(ℓ) → NO₃⁻ + 2H⁺ + 2e⁻
Reduction half equation: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(ℓ)

(Oxidation half equation)×3 + (Reduction half equation)×1, and cancel 6e⁻, 3H₂O(ℓ) and 6H⁺(aq) on the both sides. The overall redox equation is:
NO₂⁻(aq) + Cr₂O₇²⁻(aq) + 8H⁺(aq) → 2Cr³⁺(aq) + NO₃⁻ + 4H₂O(ℓ)

Fe^2+ = Fe^3+ + e^-  (1st half eq'n)
NO2^-  + 2H^+ =  NO + H2O  (unbalanced 2nd half equ'n) 

For the 2nd half eq'n ,we note that the nitrite ion loses an oxygen.  This oygen is balanced on the RHS by water. To balance the hydrogens(water) ,we write two hydrogen ions on the LHS. 
We now adjust the charges. We note that there is one negative charge (NO2^-) and two positive charges (2H^+) , so insert an 'e^-' (electron( on gthe LHS. 
Hence the 2nd half equation is 
NO2^- + 2H^+ + e^- = NO + H2O 
Since there is only one electron in each half eq'n we add the two half equ'ns. 
Hence 
  Fe^2+ = Fe^3+ + e^-  
NO2^- + 2H^+ + e^- = NO + H2O 
Adding 
NO2^-(aq) + Fe ^2+(aq) + 2H^+(aq) = Fe^3+(aq) + NO(g) + H2O(l)   

Note: the charges balance, as do the numbers of each element . 

I will leave you to do the middle one , otherwise you will not learn the method as given above. 
However, I show the similar method for the last reaction. #
NO2−(aq)+Cr2O72−(aq)→Cr3+(aq)+NO−3(aq)
 #1 
NO2^- +H2O = NO3^- + 2H^+ +2e^- 
#2 
Cr2O7^2- + 14H^+ + 4e^- = 2Cr^3+ + 7H2O

Multiply #1 by '2' to balance the electrons 
2NO2^- + 2H2O = 2NO3^- + 4H^+ + 4e^- 
Then add the two equ'ns . to eliminate the 'e^-' and to adjust the water and the acid. 
Hence 
Cr2O7^2- + 8H^+ + 2NO2^- = 2Cr^3+ + 5H2O + 2NO3^-