Find all $x$ and $y$ which are rational multiples of $\pi$ with $0<x<y<\frac{\pi}{2}$ and $\tan x+\tan y =2$.
1. Given the equation \(\tan x + \tan y = 2\), we start by using the identity for the tangent of a sum:
   \[
   \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}
   \]
   Substituting \(\tan x + \tan y = 2\), we get:
   \[
   \tan(x + y) = \frac{2}{1 - \tan x \tan y}
   \]

2. Let \(a = \tan x\) and \(b = \tan y\). Then \(a + b = 2\). We also know:
   \[
   \tan(x + y) = \frac{2}{1 - ab}
   \]
   Since \(x\) and \(y\) are rational multiples of \(\pi\), we need to find \(a\) and \(b\) such that \(ab\) is rational.

3. Using the identity for \(\sin(x + y)\) and \(\cos(x + y)\):
   \[
   \sin(x + y) = \sin x \cos y + \cos x \sin y
   \]
   \[
   \cos(x + y) = \cos x \cos y - \sin x \sin y
   \]
   We can express \(\sin x\) and \(\cos x\) in terms of \(\tan x\):
   \[
   \sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}}, \quad \cos x = \frac{1}{\sqrt{1 + \tan^2 x}}
   \]
   Similarly for \(y\):
   \[
   \sin y = \frac{\tan y}{\sqrt{1 + \tan^2 y}}, \quad \cos y = \frac{1}{\sqrt{1 + \tan^2 y}}
   \]

4. Substituting these into the sum identities, we get:
   \[
   \sin(x + y) = \frac{\tan x}{\sqrt{1 + \tan^2 x}} \cdot \frac{1}{\sqrt{1 + \tan^2 y}} + \frac{1}{\sqrt{1 + \tan^2 x}} \cdot \frac{\tan y}{\sqrt{1 + \tan^2 y}}
   \]
   \[
   \cos(x + y) = \frac{1}{\sqrt{1 + \tan^2 x}} \cdot \frac{1}{\sqrt{1 + \tan^2 y}} - \frac{\tan x}{\sqrt{1 + \tan^2 x}} \cdot \frac{\tan y}{\sqrt{1 + \tan^2 y}}
   \]

5. Simplifying these expressions, we get:
   \[
   \sin(x + y) = \frac{\tan x + \tan y}{\sqrt{(1 + \tan^2 x)(1 + \tan^2 y)}}
   \]
   \[
   \cos(x + y) = \frac{1 - \tan x \tan y}{\sqrt{(1 + \tan^2 x)(1 + \tan^2 y)}}
   \]

6. Given \(\tan x + \tan y = 2\) and substituting \(a = \tan x\) and \(b = \tan y\), we have:
   \[
   \sin(x + y) = \frac{2}{\sqrt{(1 + a^2)(1 + b^2)}}
   \]
   \[
   \cos(x + y) = \frac{1 - ab}{\sqrt{(1 + a^2)(1 + b^2)}}
   \]

7. Since \(a + b = 2\), we need to find \(a\) and \(b\) such that \(ab\) is rational. Let \(a = b = 1\), then:
   \[
   \tan x = 1, \quad \tan y = 1
   \]
   This implies:
   \[
   x = \frac{\pi}{4}, \quad y = \frac{\pi}{4}
   \]
   However, this does not satisfy \(0 < x < y < \frac{\pi}{2}\).

8. Therefore, we need to find other rational multiples of \(\pi\) that satisfy the given conditions. Since \(a\) and \(b\) must be rational and \(a + b = 2\), the only possible values are \(a = 1\) and \(b = 1\), which leads to a contradiction.

9. Hence, there are no rational multiples of \(\pi\) that satisfy the given conditions.

The final answer is \(\boxed{\text{None}}\)