To find the area bounded by the angle bisectors of the given lines, we need to first find the intersection points of the lines and then find the angle bisectors passing through those points. Finally, we can calculate the area bounded by these angle bisectors.

1. Finding the intersection points:
The given lines are:
1) x^2 + y^2 - 2y = 1
2) x + y = 3

Simplifying the first equation, we get:
x^2 + y^2 - 2y - 1 = 0
Rearranging, we have:
x^2 + (y^2 - 2y + 1) - 2 = 0
x^2 + (y - 1)^2 - 2 = 0

Comparing this equation with the standard form of a circle, we have:
(x - 0)^2 + (y - 1)^2 = √2^2

This represents a circle with center (0, 1) and radius √2.

Substituting the value of y from the second equation into the equation of the circle, we get:
(x + (3 - x))^2 + (3 - x - 1)^2 = √2^2
Simplifying, we have:
(2x - 2)^2 + (2 - x)^2 = 2

Expanding and simplifying further, we get:
4x^2 - 8x + 4 + x^2 - 4x + 4 = 2
5x^2 - 12x + 6 = 0

Using the quadratic formula, we can solve for x:
x = [-(-12) ± √((-12)^2 - 4(5)(6))]/(2(5))
x = [12 ± √(144 - 120)]/10
x = [12 ± √24]/10
x = [12 ± 2√6]/10
x = (6 ± √6)/5

Substituting these values of x back into the second equation, we can find the corresponding y-values:
For x = (6 + √6)/5:
(6 + √6)/5 + y = 3
y = 3 - (6 + √6)/5

For x = (6 - √6)/5:
(6 - √6)/5 + y = 3
y = 3 - (6 - √6)/5

So, the two intersection points are:
Point A: [x1, y1] = [(6 + √6)/5, 3 - (6 + √6)/5]
Point B: [x2, y2] = [(6 - √6)/5, 3 - (6 - √6)/5]

2. Finding the angle bisectors:
The angle bisectors pass through the intersection points A and B. Let's find the equations of these angle bisectors.

The angle bisector of line 1 is perpendicular to line 1. The slope of line 1 is given by the coefficient of x:
m1 = 1

Therefore, the slope of the angle bisector of line 1 is:
m1' = -1/m1 = -1