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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.

This problem involves solving logarithmic expressions with bases 2 and 3. Learn how to simplify expressions such as (log2 2)^2 and (log3 9)^2 using logarithmic identities. Suitable for students in Grades 9-11.

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