Let me process the mathematical problems visible in the image and solve them step by step.Here are the problems extracted from the image and their solutions:

1. $\log_4{2.56} + \log_4{6.25}$:

Using the property of logarithms:
$\log_a{x} + \log_a{y} = \log_a{(x \cdot y)}$

We combine: $\log_4{(2.56 \cdot 6.25)}$

Calculate $2.56 \cdot 6.25 = 16$: $\log_4{16}$

Rewrite $16 = 4^2$, so: $\log_4{4^2} = 2$

Answer: $2$

2. $\sqrt{\log_{\sqrt{3}}{9}}$:

First, simplify $\log_{\sqrt{3}}{9}$. Rewrite $9 = (\sqrt{3})^4$, so: $\log_{\sqrt{3}}{9} = \log_{\sqrt{3}}{(\sqrt{3})^4} = 4$

Now take the square root: $\sqrt{4} = 2$

Answer: $2$

3. $\log_{0.04}{\left( 5^{4} \cdot 0.2 \right)}$:

Simplify the argument first: $5^4 = 625, \quad 625 \cdot 0.2 = 125$

Thus: $\log_{0.04}{125}$

Rewrite $0.04 = 1/25$ and $125 = 5^3$. Using the base change: $\log_{1/25}{125} = -\log_{25}{125}$

Rewrite $125 = 25^{3/2}$: $\log_{25}{125} = \log_{25}{(25^{3/2})} = 3/2$

Finally: $-\log_{25}{125} = -\frac{3}{2}$

Answer: $-\frac{3}{2}$

4. $\log_{25}{\log_{27}{(3 \sqrt[5]{3})}}$:

First, simplify $\log_{27}{(3 \sqrt[5]{3})}$. Rewrite: $3 \sqrt[5]{3} = 3^{1 + \frac{1}{5}} = 3^{\frac{6}{5}}$

Thus: $\log_{27}{(3 \sqrt[5]{3})} = \log_{27}{(3^{6/5})}$

Since $27 = 3^3$, we have: $\log_{27}{(3^{6/5})} = \frac{6/5}{3} = \frac{2}{5}$

Now for $\log_{25}{\frac{2}{5}}$: Since no simplification is possible, the final answer is left in terms of: $\log_{25}{\frac{2}{5}}$

Answer: $\log_{25}{\frac{2}{5}}$

5. $\log_a{\sqrt{\frac{a^3}{b}}}$, given $b^5 = a^4$:

Simplify the logarithmic argument: $\sqrt{\frac{a^3}{b}} = \left( \frac{a^3}{b} \right)^{1/2} = \frac{a^{3/2}}{b^{1/2}}$

Using the logarithm property: $\log_a{\frac{a^{3/2}}{b^{1/2}}} = \log_a{a^{3/2}} - \log_a{b^{1/2}}$

Simplify each term: $\log_a{a^{3/2}} = \frac{3}{2}, \quad \log_a{b^{1/2}} = \frac{1}{2} \log_a{b}$

From $b^5 = a^4$, rewrite $b = a^{4/5}$: $\log_a{b} = \log_a{(a^{4/5})} = \frac{4}{5}$

Thus: $\log_a{b^{1/2}} = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5}$

Finally: $\frac{3}{2} - \frac{2}{5} = \frac{15}{10} - \frac{4}{10} = \frac{11}{10}$

Answer: $\frac{11}{10}$

Do you have any questions or need further clarifications?

Expanding Questions:

1. Can you explain more about the properties of logarithms used in problem 1?
2. How is the base change formula applied in logarithms (problem 3)?
3. Why is $b^5 = a^4$ important in problem 5?
4. Can we generalize logarithmic solutions with fractional bases like in problem 2?
5. How does $\sqrt[5]{x}$ simplify in logarithmic expressions?

Tip: Always simplify the argument of the logarithm before applying properties for easier calculations.