Given equation of line is (x/3) + (y/4) = 1

=> 4x + 3y = 12

=> 4x + 3y - 12 = 0 ................1

Let (0,b) is the point of the y-axis whose distance from given line is 4 unit. 

When we compare equation 1 with general form of the equation Ax+By+C = 0, we get

A = 4, B = 3, C = -12 

Now perpendicular distance of a line Ax+By+C = 0 from a point (x₁ , y₁ ) is

d = |Ax₁ + By₁ + C|/√(A²  + B² )

So  perpendicular distance of a line 4x + 3y -12 = 0 from a point (0 ,b) is

     4 = |4*0 + 3*b -12|/√(4² + 3² )

=>4 = |3b -12|/√(16 + 9)

=>4 = |3b -12|/√25

=>4 = |3b -12|/5

=>4*5 = |3b -12|

=> |3b -12| = 20

Now

      3b -12 = 20 and 3b -12 = -20

=> 3b = 20+12 and 3b = -20+12

=> 3b = 32 and 3b = -8

=> b = 32/3 and b = -8/3

So the points are (0, 32/3) and (0, -8/3)