1. Solve (a) $\frac{n}{3} x - 7 + 5 \le 6$. Simplify the left side: $\frac{n}{3} x - 2 \le 6$. Add 2 to both sides: $\frac{n}{3} x \le 8$. Multiply both sides by $\frac{3}{n}$ (assuming $n \ne 0$): $x \le \frac{24}{n}$ if $n > 0$, or $x \ge \frac{24}{n}$ if $n < 0$. 2. Solve (b) $\frac{2}{3} (x+3) \le 2 (x-4) + \frac{x}{2} + 1$. Expand both sides: $\frac{2}{3}x + 2 \le 2x - 8 + \frac{x}{2} + 1$. Simplify right side: $\frac{2}{3}x + 2 \le 2x + \frac{x}{2} - 7$. Get all terms on left: $\frac{2}{3}x - 2x - \frac{x}{2} \le -7 - 2$. Common denominator 6: $\frac{4}{6}x - \frac{12}{6}x - \frac{3}{6}x \le -9$. Combine: $\frac{4 - 12 - 3}{6} x = -\frac{11}{6} x \le -9$. Multiply both sides by $-\frac{6}{11}$ (flip inequality as negative): $x \ge \frac{54}{11}$. 3. Solve (c) $\frac{9}{4} \le \frac{1}{2} (3x + 2) - \frac{3}{4} (2x -8) < \frac{13}{4}$. Expand: Left side $\frac{1}{2}(3x + 2) = \frac{3x}{2} + 1$, right side $\frac{3}{4}(2x - 8) = \frac{3x}{2} - 6$. Expression becomes: $\frac{9}{4} \le \left(\frac{3x}{2} + 1 \right) - \left(\frac{3x}{2} - 6 \right) < \frac{13}{4}$. Simplify middle: $\frac{3x}{2} + 1 - \frac{3x}{2} + 6 = 7$. So inequality is: $\frac{9}{4} \le 7 < \frac{13}{4}$, which is false since $7 \not< 3.25$. So no solution for this compound inequality. 4. Solve (d) $-3 \le 2x + \frac{5}{3} \le 5$. Subtract $\frac{5}{3}$ from all parts: $-3 - \frac{5}{3} \le 2x \le 5 - \frac{5}{3}$. Calculate numbers: $-\frac{9}{3} - \frac{5}{3} = -\frac{14}{3}$ and $\frac{15}{3} - \frac{5}{3} = \frac{10}{3}$. So $-\frac{14}{3} \le 2x \le \frac{10}{3}$. Divide all parts by 2: $-\frac{7}{3} \le x \le \frac{5}{3}$. Answers summary: (a) Interval: $x \le \frac{24}{n}$ if $n>0$; $x \ge \frac{24}{n}$ if $n<0$. (b) Interval: $\left[ \frac{54}{11}, \infty \right)$. (c) No solution. (d) Interval: $\left[ -\frac{7}{3}, \frac{5}{3} \right]$.