To solve this system of equations, we can use substitution or elimination methods. Let's start with the first two equations:

(x-2)(y+1) = 1 x - y = 3

From equation 2, we get x = y + 3. Substituting this into equation 1:

(y+3-2)(y+1) = 1
(y+1)(y+1) = 1
y^2 + 2y + 1 = 1
y^2 + 2y = 0
y(y+2) = 0

So, y = 0 or y = -2.

If y = 0, then x = 0 + 3 = 3.
If y = -2, then x = -2 + 3 = 1.

Now, let's move on to the next set of equations:

x - y = -2xy = 15

We can substitute x = y - 2 into equation 2:

(y-2)y = 15
y^2 - 2y - 15 = 0
(y - 5)(y + 3) = 0

So, y = 5 or y = -3.

If y = 5, then x = 5 - 2 = 3.
If y = -3, then x = -3 - 2 = -5.

Finally, let's solve the last set of equations:

x^2 + y^2 = 25xy = 12

Substitute x = 3, y = 5:
3^2 + 5^2 = 9 + 25 = 34 ≠ 25

Substitute x = -5, y = -3:
(-5)^2 + (-3)^2 = 25 + 9 = 34 ≠ 25

Therefore, there seem to be no solutions that satisfy all three sets of equations simultaneously.