Let's go through each part of the question from the image you provided.

Part a:

$\left( \frac{x^2 y^2 z^3}{x^3 y^3 z^2} \right)^2$

To simplify this expression, divide the powers of similar variables in the numerator by the powers in the denominator: $\frac{x^2}{x^3} = x^{-1}, \quad \frac{y^2}{y^3} = y^{-1}, \quad \frac{z^3}{z^2} = z^1$

Thus, the expression becomes: $\left( x^{-1} y^{-1} z \right)^2 = x^{-2} y^{-2} z^2$

Part b:

$\frac{8^{\frac{3}{2}} \times 25^{\frac{1}{2}}}{16^{\frac{3}{2}} \times 9}$

Evaluate the terms: $8^{\frac{3}{2}} = (2^3)^{\frac{3}{2}} = 2^4 = 16, \quad 25^{\frac{1}{2}} = \sqrt{25} = 5, \quad 16^{\frac{3}{2}} = (2^4)^{\frac{3}{2}} = 2^6 = 64$

The expression simplifies to: $\frac{16 \times 5}{64 \times 9} = \frac{80}{576} = \frac{5}{36}$

Part c:

$\frac{16^{\frac{1}{2}}}{\sqrt{7} - 2}$

Simplify the numerator: $16^{\frac{1}{2}} = \sqrt{16} = 4$

So, the expression becomes: $\frac{4}{\sqrt{7} - 2}$

To simplify further, rationalize the denominator by multiplying by $\frac{\sqrt{7} + 2}{\sqrt{7} + 2}$: $\frac{4}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{4(\sqrt{7} + 2)}{(\sqrt{7})^2 - 2^2} = \frac{4(\sqrt{7} + 2)}{7 - 4} = \frac{4(\sqrt{7} + 2)}{3}$

Thus, the simplified result is: $\frac{4(\sqrt{7} + 2)}{3}$

Would you like more details or have any questions about the solutions?

Here are some additional related questions to consider:

1. How do you handle powers with negative exponents?
2. What is the general rule for rationalizing a denominator involving square roots?
3. What is the relationship between square roots and exponents in terms of fractional powers?
4. How do you simplify complex fractions?
5. What strategies are useful when simplifying algebraic expressions involving multiple variables?

Tip: When simplifying expressions with exponents, remember to apply the laws of exponents carefully, especially when handling negative and fractional powers.