Let's work on solving the problems sequentially as requested. I'll start with the questions in order, providing detailed explanations suitable for an 11th-grade level. Here’s the breakdown:

1a.

Question: The point $A(a, b, a^2 b)$ is in the third quadrant. In which quadrant does $B(a, b-a)$ lie?

Solution:

• In the third quadrant, both $a$ and $b$ are negative ($a < 0, b < 0$).
• The coordinates of $B(a, b-a)$:
  • $a$: Still negative ($a < 0$).
  • $b-a$: Since $b < 0$ and $a < 0$, the subtraction makes $b-a$ more negative. Hence $b-a < 0$.

Thus, $B(a, b-a)$ also lies in the third quadrant.

1b.

Question: Given $K(3, 2m-6)$ and $L(m+2, -2)$, find the integer values of $m$ such that $K$ and $L$ are in the same quadrant.

Solution:

1. For $K(3, 2m-6)$:

   • $3 > 0$ (always in the right half-plane).
   • $2m-6 > 0 \Rightarrow m > 3$ (must satisfy for $K$ to stay in the top-right quadrant).

2. For $L(m+2, -2)$:

   • $m+2 > 0 \Rightarrow m > -2$ (for $x > 0$).
   • $-2 < 0$ (always in the bottom half-plane).

Thus, $K$ and $L$ can never be in the same quadrant because $K$’s $y$-coordinate is positive while $L$’s $y$-coordinate is negative for all $m$.

2.

Question: The point $A(8-m, m-3)$ lies on the x-axis, and $B(n+2, 2m-1)$ lies on the y-axis. If $C(8-m, p-1)$ is equidistant from the axes, find $m+n+p$.

Solution:

1. $A(8-m, m-3)$:

   • On the x-axis, $y=0$: $m-3 = 0 \Rightarrow m = 3$.

2. $B(n+2, 2m-1)$:

   • On the y-axis, $x=0$: $n+2 = 0 \Rightarrow n = -2$.

3. $C(8-m, p-1)$:

   • Equidistant from the axes:
     $|8-m| = |p-1|$.
     Substitute $m = 3$:
     $|8-3| = |p-1| \Rightarrow 5 = |p-1|$.
     Solve for $p$: $p = 6$ or $p = -4$.

Thus, $m+n+p = 3 + (-2) + 6 = 7$ or $3 + (-2) + (-4) = -3$.

3.

Question: Point $A(a+1, 2b-3)$ is 7 units from the x-axis. Find the maximum $y$-distance of $B(3b-5, a+3)$ from the y-axis.

Solution:

1. For $A(a+1, 2b-3)$:

   • Distance from the x-axis is $|2b-3| = 7$.
     Solve:
     $2b-3 = 7$ or $2b-3 = -7$.
     $2b = 10 \Rightarrow b = 5$ or $2b = -4 \Rightarrow b = -2$.

2. For $B(3b-5, a+3)$:

   • Substitute $b=5$: $x=3(5)-5=10$.
     Distance to y-axis: $|x|=|10|=10$.
   • Substitute $b=-2$: $x=3(-2)-5=-11$.
     Distance to y-axis: $|x|=|-11|=11$.

Maximum distance is 11 units.

4.

Question: Point $A(a+2, 2a+7)$ is equidistant from the axes. What values can $a$ take?

Solution:

• Equidistant from axes: $|a+2| = |2a+7|$.
  Solve:
  1. $a+2 = 2a+7$:
     $a = -5$.
  2. $a+2 = -(2a+7)$:
     $a+2 = -2a-7 \Rightarrow 3a = -9 \Rightarrow a = -3$.

Thus, $a = -5$ or $a = -3$.

I’ll continue solving the remaining problems if you'd like! Would you like to explore specific problems further or continue sequentially?

Next Steps:

1. Should I move on to Problem 5?
2. Do you want detailed explanations for any step above?
3. Would you like a summary of concepts involved here?
4. Do you need clarifications on distance or quadrants?
5. Should I prioritize geometry-related or algebra-heavy problems?

Tip: To check quadrant-related questions, always evaluate the sign of both $x$- and $y$-coordinates systematically.