In a scalene triangle one angle is exactly two times as big as another one and some angle in this triangle is $36^o$. Find all possibilities, how big the angles of this triangle can be.
Given a scalene triangle where one angle is exactly twice another angle and one of the angles is \(36^\circ\), we need to find all possible sets of angles for this triangle.

Let the angles of the triangle be \(x\), \(2x\), and \(y\). We know that the sum of the angles in any triangle is \(180^\circ\).

### Case 1: The known angle \(36^\circ\) is neither \(x\) nor \(2x\)
1. Let \(y = 36^\circ\).
2. Then, the equation for the sum of the angles is:
   \[
   x + 2x + 36^\circ = 180^\circ
   \]
3. Simplify the equation:
   \[
   3x + 36^\circ = 180^\circ
   \]
4. Solve for \(x\):
   \[
   3x = 180^\circ - 36^\circ
   \]
   \[
   3x = 144^\circ
   \]
   \[
   x = 48^\circ
   \]
5. Therefore, the angles are:
   \[
   x = 48^\circ, \quad 2x = 96^\circ, \quad y = 36^\circ
   \]
   So, the angles in this case are \((36^\circ, 48^\circ, 96^\circ)\).

### Case 2: The known angle \(36^\circ\) is \(2x\)
1. Let \(2x = 36^\circ\).
2. Solve for \(x\):
   \[
   x = \frac{36^\circ}{2} = 18^\circ
   \]
3. The remaining angle is:
   \[
   180^\circ - x - 2x = 180^\circ - 18^\circ - 36^\circ = 126^\circ
   \]
4. Therefore, the angles are:
   \[
   x = 18^\circ, \quad 2x = 36^\circ, \quad y = 126^\circ
   \]
   So, the angles in this case are \((18^\circ, 36^\circ, 126^\circ)\).

### Case 3: The known angle \(36^\circ\) is \(x\)
1. Let \(x = 36^\circ\).
2. Then, \(2x = 72^\circ\).
3. The remaining angle is:
   \[
   180^\circ - x - 2x = 180^\circ - 36^\circ - 72^\circ = 72^\circ
   \]
4. This would imply that the triangle has two angles of \(72^\circ\), making it an isosceles triangle, which contradicts the given condition that the triangle is scalene. Therefore, this case is not possible.

The final answer is \( \boxed{ (36^\circ, 48^\circ, 96^\circ) } \) and \((18^\circ, 36^\circ, 126^\circ)\).