This problem involves logarithmic expressions and requires careful substitution and calculation. Let's go through the steps for solving the equations provided.

First Section:

We are given: $a = 2, M = 2 \text{ in the first case.}$

We see expressions involving $(\log_2 2)^2$. Recall that: $\log_b b = 1 \quad \text{(where the base and the argument are the same)}.$ Thus, $\log_2 2 = 1$, and: $(\log_2 2)^2 = 1^2 = 1.$ This confirms the boxed "1" in the first part.

Second Section:

Now, $a = 3$ and $M = \frac{1}{9}$. The expression becomes: $\left(\log_2 2\right)^2 \quad \text{again results in} \quad 1.$

The next part deals with the expression involving $\log_3 9$. Using the logarithmic identity: $\log_b (b^n) = n.$ Thus, $\log_3 9 = \log_3 3^2 = 2$. So: $\left(\log_3 9\right)^2 = 2^2 = 4.$ This results in a boxed "4."

Third Section:

For the next logarithmic expression: $\log_3 \left(\frac{1}{9}\right) = \log_3 \left(9^{-1}\right) = -\log_3 9 = -2.$ Therefore: $\left(\log_3 \left(\frac{1}{9}\right)\right)^2 = (-2)^2 = 4.$ This matches the boxed "4" in the equation.

The next boxed value is confirmed as $3$, following similar reasoning.

If you need further details or clarifications on any of these steps, feel free to ask!

5 Related Questions:

1. What is the general rule for logarithms when dealing with fractions as the argument?
2. How do we handle logarithmic identities when the argument is raised to a power?
3. Why does $\log_b b$ always equal 1?
4. Can you apply these logarithmic principles to bases other than 2 or 3?
5. How would the calculations change if the base of the logarithms were irrational numbers?

Tip:

Always apply logarithmic properties step by step, particularly using the identity $\log_b (x^n) = n \log_b x$, as it simplifies many expressions easily.