[b]a)[/b] Solve the system
$$  \left\{\begin{matrix} 3^y-4^x=11\\ \log_4{x} +\log_3 y =3/2\end{matrix}\right. $$

[b]b)[/b] Solve the equation $ \quad 9^{\log_5 (x-2)} -5^{\log_9 (x+2)} = 4. $
**Part (a):**

We are given the system of equations:
\[
\begin{cases}
3^y - 4^x = 11 \\
\log_4{x} + \log_3{y} = \frac{3}{2}
\end{cases}
\]

1. First, let's solve the second equation for \( x \) and \( y \):
   \[
   \log_4{x} + \log_3{y} = \frac{3}{2}
   \]
   Using the change of base formula, we can rewrite the logarithms:
   \[
   \frac{\log{x}}{\log{4}} + \frac{\log{y}}{\log{3}} = \frac{3}{2}
   \]
   Since \(\log{4} = 2\log{2}\), we have:
   \[
   \frac{\log{x}}{2\log{2}} + \frac{\log{y}}{\log{3}} = \frac{3}{2}
   \]
   Multiplying through by \(2\log{2}\log{3}\) to clear the denominators:
   \[
   \log{3} \log{x} + 2\log{2} \log{y} = 3\log{2}\log{3}
   \]

2. Now, let's test \( x = 2 \) and see if it satisfies the second equation:
   \[
   \log_4{2} + \log_3{y} = \frac{3}{2}
   \]
   Since \(\log_4{2} = \frac{1}{2}\), we have:
   \[
   \frac{1}{2} + \log_3{y} = \frac{3}{2}
   \]
   Solving for \(\log_3{y}\):
   \[
   \log_3{y} = \frac{3}{2} - \frac{1}{2} = 1
   \]
   Therefore, \( y = 3 \).

3. Now, we need to check if \( x = 2 \) and \( y = 3 \) satisfy the first equation:
   \[
   3^y - 4^x = 11
   \]
   Substituting \( y = 3 \) and \( x = 2 \):
   \[
   3^3 - 4^2 = 27 - 16 = 11
   \]
   This is true, so \( x = 2 \) and \( y = 3 \) is a solution.

Since the left-hand side of the second equation is monotonically increasing and the right-hand side is constant, there is at most one solution. Therefore, \( x = 2 \) and \( y = 3 \) is the unique solution.

**Part (b):**

We are given the equation:
\[
9^{\log_5 (x-2)} - 5^{\log_9 (x+2)} = 4
\]

1. Rewrite the equation using properties of logarithms and exponents:
   \[
   (x-2)^{\log_5 9} - (x+2)^{\log_9 5} = 4
   \]
   Note that \(\log_5 9 = \frac{\log 9}{\log 5}\) and \(\log_9 5 = \frac{\log 5}{\log 9}\).

2. Let's test \( x = 7 \) and see if it satisfies the equation:
   \[
   (7-2)^{\log_5 9} - (7+2)^{\log_9 5} = 4
   \]
   Simplifying:
   \[
   5^{\log_5 9} - 9^{\log_9 5} = 4
   \]
   Since \( 5^{\log_5 9} = 9 \) and \( 9^{\log_9 5} = 5 \), we have:
   \[
   9 - 5 = 4
   \]
   This is true, so \( x = 7 \) is a solution.

Since the left-hand side of the equation is monotonically increasing and the right-hand side is constant, there is at most one solution. Therefore, \( x = 7 \) is the unique solution.

The final answer is \( \boxed{ (x, y) = (2, 3) } \) for part (a) and \( x = 7 \) for part (b).