Solve in the interval $ (2,\infty ) $ the following equation:
$$ 1=\cos\left( \pi\log_3 (x+6)\right)\cdot\cos\left( \pi\log_3 (x-2)\right) . $$
1. Given the equation:
   \[
   1 = \cos\left( \pi \log_3 (x+6) \right) \cdot \cos\left( \pi \log_3 (x-2) \right)
   \]

2. We know that the product of two cosine functions equals 1 if and only if each cosine function is either 1 or -1, and their product is 1. Therefore, we need:
   \[
   \cos\left( \pi \log_3 (x+6) \right) = \pm 1 \quad \text{and} \quad \cos\left( \pi \log_3 (x-2) \right) = \pm 1
   \]

3. The cosine function equals 1 or -1 at integer multiples of $\pi$:
   \[
   \cos(\theta) = 1 \quad \text{if} \quad \theta = 2k\pi \quad \text{for} \quad k \in \mathbb{Z}
   \]
   \[
   \cos(\theta) = -1 \quad \text{if} \quad \theta = (2k+1)\pi \quad \text{for} \quad k \in \mathbb{Z}
   \]

4. Therefore, we need:
   \[
   \pi \log_3 (x+6) = 2k\pi \quad \text{or} \quad \pi \log_3 (x+6) = (2k+1)\pi
   \]
   \[
   \pi \log_3 (x-2) = 2m\pi \quad \text{or} \quad \pi \log_3 (x-2) = (2m+1)\pi
   \]
   for some integers $k$ and $m$.

5. Simplifying these equations, we get:
   \[
   \log_3 (x+6) = 2k \quad \text{or} \quad \log_3 (x+6) = 2k+1
   \]
   \[
   \log_3 (x-2) = 2m \quad \text{or} \quad \log_3 (x-2) = 2m+1
   \]

6. Converting the logarithmic equations to exponential form, we have:
   \[
   x+6 = 3^{2k} \quad \text{or} \quad x+6 = 3^{2k+1}
   \]
   \[
   x-2 = 3^{2m} \quad \text{or} \quad x-2 = 3^{2m+1}
   \]

7. Solving for $x$ in each case:
   \[
   x = 3^{2k} - 6 \quad \text{or} \quad x = 3^{2k+1} - 6
   \]
   \[
   x = 3^{2m} + 2 \quad \text{or} \quad x = 3^{2m+1} + 2
   \]

8. We need to find common values of $x$ that satisfy both equations. Let's test $x = 3$:
   \[
   \log_3 (3+6) = \log_3 9 = 2 \quad \text{and} \quad \log_3 (3-2) = \log_3 1 = 0
   \]
   \[
   \cos(\pi \cdot 2) = \cos(2\pi) = 1 \quad \text{and} \quad \cos(\pi \cdot 0) = \cos(0) = 1
   \]
   \[
   1 \cdot 1 = 1
   \]
   Thus, $x = 3$ is a solution.

9. To check if there are any other solutions, we note that $\cos(\pi \log_3 (x+6))$ and $\cos(\pi \log_3 (x-2))$ are periodic functions with period $2\pi$. However, since the product of two cosine functions equals 1 only when both are 1 or both are -1, and given the constraints of the logarithmic functions, $x = 3$ is the only solution in the interval $(2, \infty)$.

The final answer is $\boxed{x = 3}$