To find which point is not a solution to the given system of linear inequalities:

1. $y < 2x + 2$
2. $y \leq -\frac{5}{2}x - 5$

We need to check each provided point to see if they satisfy both inequalities.

Point (-4, -8)

1. For $y < 2x + 2$: $-8 < 2(-4) + 2$ $-8 < -8 + 2$ $-8 < -6 \quad (\text{True})$

2. For $y \leq -\frac{5}{2}x - 5$: $-8 \leq -\frac{5}{2}(-4) - 5$ $-8 \leq 10 - 5$ $-8 \leq 5 \quad (\text{True})$

Point (-8, -1)

1. For $y < 2x + 2$: $-1 < 2(-8) + 2$ $-1 < -16 + 2$ $-1 < -14 \quad (\text{False})$

2. For $y \leq -\frac{5}{2}x - 5$: $-1 \leq -\frac{5}{2}(-8) - 5$ $-1 \leq 20 - 5$ $-1 \leq 15 \quad (\text{True})$

Point (-2, -4)

1. For $y < 2x + 2$: $-4 < 2(-2) + 2$ $-4 < -4 + 2$ $-4 < -2 \quad (\text{True})$

2. For $y \leq -\frac{5}{2}x - 5$: $-4 \leq -\frac{5}{2}(-2) - 5$ $-4 \leq 5 - 5$ $-4 \leq 0 \quad (\text{True})$

Point (-2, -8)

1. For $y < 2x + 2$: $-8 < 2(-2) + 2$ $-8 < -4 + 2$ $-8 < -2 \quad (\text{True})$

2. For $y \leq -\frac{5}{2}x - 5$: $-8 \leq -\frac{5}{2}(-2) - 5$ $-8 \leq 5 - 5$ $-8 \leq 0 \quad (\text{True})$

Conclusion

The point $(-8, -1)$ does not satisfy the first inequality $y < 2x + 2$ and therefore is not a solution to the system of inequalities.