Let $ x,y\in\mathbb{R}$ , and $ x,y  \in  $  $ \left(0,\frac{\pi}{2}\right)  $, and $  m \in \left(2,+\infty\right) $ such that $ \tan x * \tan y = m $ . Find the minimum value of the expression $ E(x,y) = \cos x + \cos y $.
Given \( x, y \in \left(0, \frac{\pi}{2}\right) \) and \( m \in \left(2, +\infty\right) \) such that \( \tan x \cdot \tan y = m \), we need to find the minimum value of the expression \( E(x, y) = \cos x + \cos y \).

1. **Expressing the given condition in terms of trigonometric identities:**
   \[
   \tan x \cdot \tan y = m \implies \frac{\sin x}{\cos x} \cdot \frac{\sin y}{\cos y} = m \implies \frac{\sin x \sin y}{\cos x \cos y} = m
   \]
   This implies:
   \[
   \sin x \sin y = m \cos x \cos y
   \]

2. **Using the product-to-sum identities:**
   \[
   \sin x \sin y = \frac{1}{2} [\cos(x-y) - \cos(x+y)]
   \]
   \[
   \cos x \cos y = \frac{1}{2} [\cos(x-y) + \cos(x+y)]
   \]
   Substituting these into the given condition:
   \[
   \frac{1}{2} [\cos(x-y) - \cos(x+y)] = m \cdot \frac{1}{2} [\cos(x-y) + \cos(x+y)]
   \]
   Simplifying:
   \[
   \cos(x-y) - \cos(x+y) = m \cos(x-y) + m \cos(x+y)
   \]
   \[
   \cos(x-y) - m \cos(x-y) = m \cos(x+y) + \cos(x+y)
   \]
   \[
   (1 - m) \cos(x-y) = (m + 1) \cos(x+y)
   \]
   \[
   \frac{\cos(x-y)}{\cos(x+y)} = \frac{m+1}{m-1}
   \]

3. **Introducing new variables:**
   Let \( x - y = 2a \) and \( x + y = 2b \). Then:
   \[
   \frac{\cos(2a)}{\cos(2b)} = \frac{m+1}{m-1}
   \]
   Let \( k = \frac{\cos(2a)}{1+m} = \frac{\cos(2b)}{1-m} \). Then:
   \[
   \cos(2a) = k(1+m) \quad \text{and} \quad \cos(2b) = k(1-m)
   \]

4. **Expressing \( E(x, y) \) in terms of \( a \) and \( b \):**
   \[
   E(x, y) = \cos x + \cos y
   \]
   Using the double-angle identities:
   \[
   \cos x = \sqrt{\frac{1 + \cos(2a)}{2}} \quad \text{and} \quad \cos y = \sqrt{\frac{1 + \cos(2b)}{2}}
   \]
   Therefore:
   \[
   E(x, y) = \sqrt{\frac{1 + k(1+m)}{2}} + \sqrt{\frac{1 + k(1-m)}{2}}
   \]

5. **Finding the minimum value of \( E(x, y) \):**
   To find the minimum value, we need to analyze the behavior of the function:
   \[
   E^2 = \left( \cos x + \cos y \right)^2 = \cos^2 x + \cos^2 y + 2 \cos x \cos y
   \]
   Using the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \):
   \[
   E^2 = \frac{1 + \cos(2a)}{2} + \frac{1 + \cos(2b)}{2} + 2 \cos x \cos y
   \]
   \[
   E^2 = \frac{1 + k(1+m)}{2} + \frac{1 + k(1-m)}{2} + 2 \cos x \cos y
   \]
   Simplifying:
   \[
   E^2 = \frac{2 + k(1+m) + k(1-m)}{2} + 2 \cos x \cos y
   \]
   \[
   E^2 = 1 + k + 2 \cos x \cos y
   \]
   Since \( \cos x \cos y = \frac{1}{2} [\cos(x-y) + \cos(x+y)] \):
   \[
   E^2 = 1 + k + \cos(x-y) + \cos(x+y)
   \]
   Using the earlier result:
   \[
   E^2 = 1 + k + \frac{m+1}{m-1} + \frac{m-1}{m+1}
   \]
   Simplifying further, we find that the minimum value of \( E(x, y) \) is achieved when \( k \) is minimized.

The final answer is \(\boxed{2}\).