The table above shows the test scores of students in a class. If the average score of 10 boys was 12 years, a boy of 14 years was replaced with that of 6, find the new average of boys.

Multiply the entire inequality by 12 to eliminate the fractions.

$$12 \left( \frac{2}{3}(x-2) - \frac{1}{4}(x-5) \right) \le 12(0)$$ $$8(x-2) - 3(x-5) \le 0$$

$$(8x - 3x) + (-16 + 15) \le 0$$ $$5x - 1 \le 0$$

$$5x \le 1$$ $$x \le \frac{1}{5}$$

The solution to the inequality is $\boxed{x \le \frac{1}{5}}$.

2. Find the range of $x$ satisfying both inequalities: $2x+2 < 6$ and $7x+4 \ge 24$.

$$2x+2 < 6$$ $$2x < 6 - 2$$ $$2x < 4$$ $$x < 2$$

$$7x+4 \ge 24$$ $$7x \ge 24 - 4$$ $$7x \ge 20$$ $$x \ge \frac{20}{7}$$ $$x \ge 2.857...$$

We need $x < 2$ AND $x \ge \frac{20}{7}$.
Since $\frac{20}{7} \approx 2.86$, there is no value of $x$ that can be simultaneously less than 2 and greater than or equal to 2.86.
Therefore, there is no range of $x$ that satisfies both inequalities.

3. Find the mean deviation of the numbers 2, 3, 5, 6.

$$|2 - 4| = |-2| = 2$$ $$|3 - 4| = |-1| = 1$$ $$|5 - 4| = |1| = 1$$ $$|6 - 4| = |2| = 2$$

$$\text{Mean Deviation} = \frac{\sum |x - \bar{x}|}{N} = \frac{6}{4} = 1.5$$

The mean deviation is $\boxed{1.5}$.

4. What is the median and semi-interquartile range of the distribution below?

Scores ($x$)	Frequency ($f$)
1	8
2	11
3	13
4	14
5	5
6	4
7	5
8	8

The median is the value at the $\frac{N+1}{2}$-th position.

$$\text{Position of } Q_2 = \frac{68+1}{2} = \frac{69}{2} = 34.5\text{-th position}$$

From the cumulative frequency, the 34.5-th value falls within the score of 4 (since $cf=32$ for score 3 and $cf=46$ for score 4).
So, the median ($Q_2$) = 4.

The first quartile is the value at the $\frac{N+1}{4}$-th position.

$$\text{Position of } Q_1 = \frac{68+1}{4} = \frac{69}{4} = 17.25\text{-th position}$$

From the cumulative frequency, the 17.25-th value falls within the score of 2 (since $cf=8$ for score 1 and $cf=19$ for score 2).
So, $Q_1$ = 2.

The third quartile is the value at the $\frac{3(N+1)}{4}$-th position.

$$\text{Position of } Q_3 = \frac{3(68+1)}{4} = \frac{3 \times 69}{4} = \frac{207}{4} = 51.75\text{-th position}$$

From the cumulative frequency, the 51.75-th value falls within the score of 6 (since $cf=51$ for score 5 and $cf=55$ for score 6).
So, $Q_3$ = 6.

$$\text{SIQR} = \frac{Q_3 - Q_1}{2} = \frac{6 - 2}{2} = \frac{4}{2} = 2$$

The median is $\boxed{4}$ and the semi-interquartile range is $\boxed{2}$.

5. The average age of 10 boys was 12 years. A boy of 14 years was replaced with that of 6 years. Find the new average age of boys.

$$\text{Total initial age} = \text{Average age} \times \text{Number of boys}$$ $$\text{Total initial age} = 12 \text{ years} \times 10 = 120 \text{ years}$$

$$\text{Change in total age} = \text{Age of new boy} - \text{Age of old boy}$$ $$\text{Change in total age} = 6 \text{ years} - 14 \text{ years} = -8 \text{ years}$$

$$\text{New total age} = \text{Total initial age} + \text{Change in total age}$$ $$\text{New total age} = 120 \text{ years} - 8 \text{ years} = 112 \text{ years}$$

The number of boys remains 10.

$$\text{New average age} = \frac{\text{New total age}}{\text{Number of boys}}$$ $$\text{New average age} = \frac{112 \text{ years}}{10} = 11.2 \text{ years}$$

The new average age of the boys is $\boxed{11.2 \text{ years}}$.