Compute
    \[
        \left(\frac{4-\log_{36} 4 - \log_6 {18}}{\log_4 3} \right) \cdot \left( \log_8 {27} + \log_2 9 \right).
    \]

[i]2020 CCA Math Bonanza Team Round #4[/i]
1. Start by simplifying the expression inside the first fraction:
    \[
    \left(\frac{4 - \log_{36} 4 - \log_6 18}{\log_4 3}\right)
    \]

2. Use the change of base formula for logarithms:
    \[
    \log_{36} 4 = \frac{\log 4}{\log 36} = \frac{\log 4}{2 \log 6} = \frac{\log 2^2}{2 \log 6} = \frac{2 \log 2}{2 \log 6} = \frac{\log 2}{\log 6}
    \]
    \[
    \log_6 18 = \log_6 (6 \cdot 3) = \log_6 6 + \log_6 3 = 1 + \log_6 3
    \]

3. Substitute these back into the expression:
    \[
    4 - \log_{36} 4 - \log_6 18 = 4 - \frac{\log 2}{\log 6} - (1 + \log_6 3)
    \]
    \[
    = 4 - \frac{\log 2}{\log 6} - 1 - \log_6 3
    \]
    \[
    = 3 - \frac{\log 2}{\log 6} - \log_6 3
    \]

4. Use the change of base formula again:
    \[
    \log_6 2 = \frac{\log 2}{\log 6}, \quad \log_6 3 = \frac{\log 3}{\log 6}
    \]
    \[
    3 - \log_6 2 - \log_6 3 = 3 - \frac{\log 2}{\log 6} - \frac{\log 3}{\log 6}
    \]
    \[
    = 3 - \frac{\log 2 + \log 3}{\log 6} = 3 - \frac{\log 6}{\log 6} = 3 - 1 = 2
    \]

5. Now, the expression simplifies to:
    \[
    \left(\frac{2}{\log_4 3}\right)
    \]

6. Next, simplify the second part of the expression:
    \[
    \log_8 27 + \log_2 9
    \]

7. Use the change of base formula:
    \[
    \log_8 27 = \frac{\log 27}{\log 8} = \frac{\log 3^3}{\log 2^3} = \frac{3 \log 3}{3 \log 2} = \frac{\log 3}{\log 2}
    \]
    \[
    \log_2 9 = \log_2 3^2 = 2 \log_2 3
    \]

8. Combine these results:
    \[
    \log_8 27 + \log_2 9 = \frac{\log 3}{\log 2} + 2 \log_2 3
    \]
    \[
    = \log_2 3 + 2 \log_2 3 = 3 \log_2 3
    \]

9. Now, the entire expression becomes:
    \[
    \left(\frac{2}{\log_4 3}\right) \cdot (3 \log_2 3)
    \]

10. Use the change of base formula for \(\log_4 3\):
    \[
    \log_4 3 = \frac{\log 3}{\log 4} = \frac{\log 3}{2 \log 2}
    \]

11. Substitute this back into the expression:
    \[
    \left(\frac{2}{\frac{\log 3}{2 \log 2}}\right) \cdot (3 \log_2 3)
    \]
    \[
    = \left(\frac{2 \cdot 2 \log 2}{\log 3}\right) \cdot (3 \log_2 3)
    \]
    \[
    = \left(\frac{4 \log 2}{\log 3}\right) \cdot (3 \log_2 3)
    \]

12. Simplify further:
    \[
    = 4 \cdot 3 \cdot \frac{\log 2 \cdot \log_2 3}{\log 3}
    \]
    \[
    = 4 \cdot 3 \cdot \frac{\log 2 \cdot \frac{\log 3}{\log 2}}{\log 3}
    \]
    \[
    = 4 \cdot 3 \cdot 1 = 12
    \]

The final answer is \(\boxed{12}\)