The image contains an exercise asking to express mathematical expressions as powers. Let's break them down one by one:

a) $(-5)^{-2} \cdot 2^{-2}$

Using the property of exponents: $(-5)^{-2} = \frac{1}{(-5)^2} = \frac{1}{25}, \quad 2^{-2} = \frac{1}{2^2} = \frac{1}{4}$ Thus: $(-5)^{-2} \cdot 2^{-2} = \frac{1}{25} \cdot \frac{1}{4} = \frac{1}{100}$

b) \left(\frac{2}{3}\right)^{-3} \cdot \left(\frac{2}{3}\right)^{-2}

Using the property of exponents: $\left(\frac{2}{3}\right)^{-3} \cdot \left(\frac{2}{3}\right)^{-2} = \left(\frac{2}{3}\right)^{-3-2} = \left(\frac{2}{3}\right)^{-5}$ Then: $\left(\frac{2}{3}\right)^{-5} = \frac{1}{\left(\frac{2}{3}\right)^5} = \frac{3^5}{2^5} = \frac{243}{32}$

c) \left(\frac{2}{7}\right)^{5} \cdot \left(\frac{2}{7}\right)^{-2}

Using the same exponent rule: $\left(\frac{2}{7}\right)^{5} \cdot \left(\frac{2}{7}\right)^{-2} = \left(\frac{2}{7}\right)^{5-2} = \left(\frac{2}{7}\right)^{3}$

d) \left(\frac{4}{5}\right)^{-5} \cdot \left(\frac{4}{5}\right)^{-5}

Again applying the exponent rule: $\left(\frac{4}{5}\right)^{-5} \cdot \left(\frac{4}{5}\right)^{-5} = \left(\frac{4}{5}\right)^{-5-5} = \left(\frac{4}{5}\right)^{-10}$ Thus: $\left(\frac{4}{5}\right)^{-10} = \frac{1}{\left(\frac{4}{5}\right)^{10}} = \frac{5^{10}}{4^{10}}$

e) \left((-0.1)^2\right)^0

Any number raised to the power of 0 is 1: $\left((-0.1)^2\right)^0 = 1$

f) $(3,37)^{-3}$

Apply the negative exponent rule: $(3,37)^{-3} = \frac{1}{(3.37)^3}$

g) $0.3^2 \cdot 0.3^{-5}$

Simplify using the rule for multiplying powers: $0.3^2 \cdot 0.3^{-5} = 0.3^{2-5} = 0.3^{-3}$ Thus: $0.3^{-3} = \frac{1}{0.3^3}$

h) $(-3)^3 \cdot (-4)^3$

Calculate each separately: $(-3)^3 = -27, \quad (-4)^3 = -64$ Then: $(-3)^3 \cdot (-4)^3 = (-27) \cdot (-64) = 1728$

i) $3^7 \cdot 2.5^7$

Using the same base exponent rule: $3^7 \cdot 2.5^7 = (3 \cdot 2.5)^7 = 7.5^7$

j) $(-2.5)^{-3} \cdot (-2.5)^4$

Using exponent rules: $(-2.5)^{-3} \cdot (-2.5)^4 = (-2.5)^{-3+4} = (-2.5)^1 = -2.5$

k) $(2,4^{-2})^5$

Simplifying inside the parentheses: $2.4^{-2} = \frac{1}{2.4^2}$ Now: $\left(\frac{1}{2.4^2}\right)^5 = \frac{1}{2.4^{10}}$

l) \left(\left(2 \cdot \frac{3}{4}\right)^{-3}\right)^{-2}

First simplify the fraction: $2 \cdot \frac{3}{4} = \frac{6}{4} = 1.5$ Then apply the exponents: $(1.5)^{-3} = \frac{1}{1.5^3}$ And then: $\left(\frac{1}{1.5^3}\right)^{-2} = 1.5^6$

Let me know if you'd like further details or explanations!

Here are five related questions to expand on this:

1. What are the key rules of exponents used in these problems?
2. How do negative exponents affect the base number?
3. Why does any base raised to the power of zero equal 1?
4. How do fractional exponents relate to roots?
5. How do you simplify expressions with the same base raised to different powers?

Tip: When multiplying numbers with the same base, simply add the exponents.