How to balance a redox reaction equation if there is already a coffient in the equation?

You start by ignoring the given coefficient and balancing the equation in the usual way:
N2O4 → NO3 + NO [balanced as written]
Then you multiply everything by whatever integer makes the given coefficient come out the way it was given. In this case, multiply by 2:
2 N2O4 → 2 NO3 + 2 NO

(Although there had better be a good reason to do it, since equations with inflated coefficients like this are usually forbidden.)

Note that NO₃ does not exist. The following is just to illustrate the method of balancing chemical equations.

Assume that the balanced equation is 2 N₂O₄ → x NO₃ + y NO

Number of N atoms :
2 × 2 = x + y
x + y = 4 ...... [1]

Mumber of O atoms :
2 × 4 = 3x + y
3x + y = 8 ...... [2]

[2] - [1] :
2x = 4
x = 2

Sub. x = 2 into [1] :
2 + y = 4
y = 2

The answer is : 2 N₂O₄ → 2 NO₃ + 2 NO

(The equation can be simplified as : N₂O₄ → NO₃ + NO

You can use half-reactions.

2N2O4 ---> NO3 + NO

leads to:

N2O4 ---> NO3
N2O4 ---> NO

Balance in acidic solution:

2H2O + N2O4 ---> 2NO3 + 4H^+ + 4e-
4e- + 4H^+ + N2O4 ---> 2NO + 2H2O

Add and eliminate like items:

2N2O4 ---> 2NO3 + 2NO

Reduce:

N2O4 ---> NO3 + NO