The uploaded image contains several math problems. Let me analyze the content and solve each one step by step:

Problem [5]: Factorize the following equations.

(1) $x^2 + 2x - 8$

Factorizing: $x^2 + 2x - 8 = (x + 4)(x - 2)$ Answer: $(x + 4)(x - 2)$

(2) $x^2 + 8x + 16$

This is a perfect square trinomial: $x^2 + 8x + 16 = (x + 4)^2$ Answer: $(x + 4)^2$

(3) $x^2 - 64$

This is a difference of squares: $x^2 - 64 = (x + 8)(x - 8)$ Answer: $(x + 8)(x - 8)$

Problem [6]: The table shows the test scores of 7 students (A to G) where 70 is the base score. Their deviations are given. Find the average score.

Table:

To find the average score:

1. Add all deviations:
   $11 + 9 - 4 - 6 - 13 + 18 + 15 = 30$

2. Divide by the number of students ($7$):
   $\frac{30}{7} \approx 4.29$

3. Add this average deviation to the base score ($70$):
   $70 + 4.29 = 74.29$

Answer: The average score is approximately 74.29 points.

Problem [7]: Solve $a = 5(b + 3)$ for $b$.

Rearrange the equation: $a = 5(b + 3)$ Divide both sides by 5: $b + 3 = \frac{a}{5}$ Subtract 3 from both sides: $b = \frac{a}{5} - 3$

Answer: $b = \frac{a}{5} - 3$

Problem [8]: If $x = \frac{1}{3}$ and $y = -2$, find $9x^2y^2 - 2xy$.

Substitute $x = \frac{1}{3}$ and $y = -2$ into the expression: $9x^2y^2 - 2xy$

Step-by-step:

1. $x^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9}$
2. $y^2 = (-2)^2 = 4$
3. $x^2y^2 = \frac{1}{9} \cdot 4 = \frac{4}{9}$
4. $9x^2y^2 = 9 \cdot \frac{4}{9} = 4$
5. $xy = \frac{1}{3} \cdot (-2) = -\frac{2}{3}$
6. $-2xy = -2 \cdot -\frac{2}{3} = \frac{4}{3}$

Combine: $9x^2y^2 - 2xy = 4 + \frac{4}{3} = \frac{12}{3} + \frac{4}{3} = \frac{16}{3}$

Answer: $\frac{16}{3}$

Problem [9]: How many natural numbers $a$ satisfy $1 < \sqrt{a} < 2$?

The inequality $1 < \sqrt{a} < 2$ implies: [ 1^2 < a < 2^2 \quad