Let's work through each of the three problems step by step.

5) $(7y^4)^2$

Using the property of exponents $(a^m)^n = a^{m \cdot n}$, we apply the exponent to both the number $7$ and the variable $y^4$:

$(7y^4)^2 = 7^2 \cdot (y^4)^2 = 49 \cdot y^{4 \cdot 2} = 49y^8$

So the simplified result is: $49y^8$

7) $(2x^4 y^4)^3$

Again, using the exponent rule $(a^m)^n = a^{m \cdot n}$, we apply the exponent to each factor inside the parentheses:

$(2x^4 y^4)^3 = 2^3 \cdot (x^4)^3 \cdot (y^4)^3 = 8 \cdot x^{4 \cdot 3} \cdot y^{4 \cdot 3} = 8x^{12}y^{12}$

So the simplified result is: $8x^{12}y^{12}$

9) $(x^2 y^2)^2$

Using the same exponent rule, we apply the outer exponent $2$ to both the $x^2$ and the $y^2$:

$(x^2 y^2)^2 = (x^2)^2 \cdot (y^2)^2 = x^{2 \cdot 2} \cdot y^{2 \cdot 2} = x^4 y^4$

So the simplified result is: $x^4 y^4$

Would you like a more detailed explanation for any of these steps?

Here are some related questions you might find interesting:

1. How would you simplify $(3a^2b^5)^3$?
2. What is the result of $(5p^3q^2)^4$?
3. Can you apply the same method to simplify $(4m^5n^3)^2$?
4. What happens if one of the bases inside the parentheses is a fraction, e.g., $(\frac{1}{2}x^2)^3$?
5. How would the results change if we had negative exponents, like $(x^{-3}y^{-2})^2$?

Tip: Always remember that when raising a product to a power, you can distribute the exponent to each factor inside the parentheses!