a) Khi m = -2, ta có hệ phương trình:
-2x + 2y = -10
(1 - (-2))x + y = 10
Simplifying the second equation, we have:
3x + y = 10
Now we have the system of equations:
-2x + 2y = -10
3x + y = 10
To solve this system, we can use the method of substitution or elimination. Let's use the method of elimination:
Multiply the first equation by 3 and the second equation by 2 to eliminate the y term:
-6x + 6y = -30
6x + 2y = 20
Add the two equations together:
8y = -10
y = -10/8
y = -5/4
Substitute this value of y back into one of the original equations to solve for x:
-2x + 2(-5/4) = -10
-2x - 10/2 = -10
-2x - 5 = -10
-2x = -5
x = -5/-2
x = 5/2
So the solution to the system of equations when m = -2 is x = 5/2 and y = -5/4.

b) To find the value of m for the system to have a unique solution, we can use the determinant of the coefficient matrix. The coefficient matrix for the system is:
| m 2m |
| 1-m 1 |
The determinant of this matrix is:
det = m(1) - (2m)(1-m)
det = m - 2m + 2m^2
det = 2m^2 - m
For the system to have a unique solution, the determinant must be non-zero. So we need to find the values of m for which det ≠ 0.
Setting the determinant equal to zero and solving for m:
2m^2 - m = 0
m(2m - 1) = 0
m = 0 or m = 1/2
Therefore, the values of m for the system to have a unique solution are m = 0 and m = 1/2.