2x + 3 > 5 => 2x > 2 => x > 1.

2x + 3 > 7 => 2x > 4 => x > 2.

2x + 3 ≤ 7 => 2x ≤ 4 => x ≤ 2.

3x - 4 < 2x + 5 => x < 9.

4x - 1 > 3 => 4x > 4 => x > 1.

4x - 1 > 3x + 2 => x > 3.

4x - 1 < 3 => 4x < 4 => x < 1.

4x - 8 < 0 => 4x < 8 => x < 2.

5 - 2x ≤ 3 => -2x ≤ -2 => x ≥ 1.

5 - x ≥ 2 => -x ≥ -3 => x ≤ 3.

5x - 1 > 4 => 5x > 5 => x > 1.

5x - 7 < 3 => 5x < 10 => x < 2.

5x - 7 ≥ 3 => 5x ≥ 10 => x ≥ 2.

6x + 4 < 10 => 6x < 6 => x < 1.

6x - 4 < 2x + 8 => 4x < 12 => x < 3.

7 - 3x < 1 => -3x < -6 => x > 2.

Subtract 2 from both sides: 7x ≥ 2. Then divide by 7: x ≥ 2/7.

7x + 3 < 4x + 12 => 3x < 9 => x < 3.

Add 4 to both sides: 7x ≥ 14. Then divide by 7: x ≥ 2.

Setting f(1-) = f(1+) and f'(1-) = f'(1+) leads to a = 1 for differentiability.

Setting the two pieces equal at x = 1 gives us 3 + c = 1. Thus, c = -2.

To ensure continuity at x = 1, k(1) + 1 = 2(1) - 3, solving gives k = 2.

To ensure continuity at x = 1, we need to set the two pieces equal: 1^2 + k = 2(1) + 1. This gives k = 2.

To ensure continuity at x = 1, we need to set the two pieces equal: k + 1^2 = 2(1) + 3. This gives k + 1 = 5, so k = 4.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right. Thus, k must equal 0.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right and equal to f(2). Thus, k = 4.

Setting f(2-) = f(2+) and f'(2-) = f'(2+) leads to m = 1 for differentiability.

Setting the two pieces equal at x = 0: n^2 - 1 = 3. Solving gives n = ±2.

Setting 2(2) + 3 = p(2) + 1 gives p = 3 for continuity.

Setting 3(2) - 1 = p(2) + 4 gives p = 2.