To find the value of \(x_1 + x_2\) given the conditions in the question, we need to understand and utilize the properties of the matrices involved.

1. We are given that \(B = (I + A)^{-1}\) and \(A + C = I\). This implies that \(C = I - A\).
2. We also know that \(BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}\).
3. Since \(B = (I + A)^{-1}\) and \(C = I - A\), observe that \( (I+A)B = I \). Therefore, \( (I+A)BC = C \).
4. Substitute \( C = I - A \) into the equation: \[ (I+A)BC = C \implies (I+A) \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix} = I - A \] This suggests that the multiplication of matrices \(I + A\) and the matrix on the left side retains that property.
5. From \((I + A)B = I\), it implies that multiplying both sides by any conforming matrix will not affect the property, hence \(BC\) retains the same matrix structure even as it assures derivations in computations with \(B\) and \(C\).
6. Now, the problem gives that \(CB \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}\). Assuming the invertibility and symmetry, compute \(C\) as the left inverse of \(B\).
7. Eligible strategy if something like \(CB = I\) then simplifies the application: \[ \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix} \] Expanding, we get two equations: \[ x_1 - 5x_2 = 12, \] \[ -x_1 + 2x_2 = -6 \]
8. Solving these equations, sum the two equations: \[ (x_1 - 5x_2) + (-x_1 + 2x_2) = 12 - 6 \] Which simplifies to: \[ -3x_2 = 6 \implies x_2 = -2 \]
9. Substitute \(x_2 = -2\) back into the first equation: \[ x_1 - 5(-2) = 12 \implies x_1 + 10 = 12 \implies x_1 = 2 \]
10. Thus, the sum \(x_1 + x_2 = 2 - 2 = 0\).

Hence, the answer is 0.