Let's evaluate each logarithm step by step:

a) $\log_5 5$

By the definition of a logarithm, $\log_b b = 1$.
Thus, $\log_5 5 = 1$.

b) $\log_7 1$

By the property of logarithms, $\log_b 1 = 0$ (for any $b > 0, b \neq 1$).
Thus, $\log_7 1 = 0$.

c) $\log_2 \left(\frac{1}{4}\right)$

We rewrite $\frac{1}{4}$ as $4^{-1}$, and $4 = 2^2$, so:
$\frac{1}{4} = 2^{-2}.$ Using the property $\log_b (a^n) = n \cdot \log_b a$, we get:
$\log_2 \left(\frac{1}{4}\right) = \log_2 \left(2^{-2}\right) = -2.$

d) $\log_7 \sqrt{7}$

We rewrite $\sqrt{7}$ as $7^{1/2}$. Using $\log_b (a^n) = n \cdot \log_b a$, we get:
$\log_7 \sqrt{7} = \log_7 \left(7^{1/2}\right) = \frac{1}{2} \cdot \log_7 7.$ Since $\log_7 7 = 1$, we have:
$\log_7 \sqrt{7} = \frac{1}{2}.$

e) $\log_{\frac{2}{3}} \left(\frac{8}{27}\right)$

We rewrite $\frac{8}{27}$ as $\left(\frac{2}{3}\right)^3$. Using the property $\log_b (a^n) = n \cdot \log_b a$, we get:
$\log_{\frac{2}{3}} \left(\frac{8}{27}\right) = \log_{\frac{2}{3}} \left(\left(\frac{2}{3}\right)^3\right) = 3.$

f) $\log_2 \sqrt[3]{2}$

We rewrite $\sqrt[3]{2}$ as $2^{1/3}$. Using $\log_b (a^n) = n \cdot \log_b a$, we get:
$\log_2 \sqrt[3]{2} = \log_2 \left(2^{1/3}\right) = \frac{1}{3} \cdot \log_2 2.$ Since $\log_2 2 = 1$, we have:
$\log_2 \sqrt[3]{2} = \frac{1}{3}.$

Final Answers:

a) $1$
b) $0$
c) $-2$
d) $\frac{1}{2}$
e) $3$
f) $\frac{1}{3}$

Would you like detailed explanations for any specific part? Let me know!

Here are 5 related questions to deepen your understanding:

1. What is the value of $\log_b b^k$, and why does this property hold?
2. How can we simplify $\log_b \left(\frac{x}{y}\right)$ using logarithmic rules?
3. What happens when the base $b$ of a logarithm is a fraction?
4. How does the logarithmic property $\log_b (a^n)$ apply to roots and fractions?
5. Why is $\log_b 1$ always equal to 0, regardless of $b$?

Tip: Always look for opportunities to rewrite numbers or fractions in terms of powers or roots of the base to simplify logarithms.