The equation is given
$x^2-(m+3)x+m+2=0$.
If $x_1$ and $x_2$ are its solutions find all $m$ such that 
$\frac{x_1}{x_1+1}+\frac{x_2}{x_2+1}=\frac{13}{10}$.

1. Given the quadratic equation \(x^2 - (m+3)x + m+2 = 0\), we know from Vieta's formulas that the sum of the roots \(x_1 + x_2 = m+3\) and the product of the roots \(x_1 x_2 = m+2\).

2. We need to find \(m\) such that \(\frac{x_1}{x_1+1} + \frac{x_2}{x_2+1} = \frac{13}{10}\).

3. Combine the fractions:
   \[
   \frac{x_1}{x_1+1} + \frac{x_2}{x_2+1} = \frac{x_1(x_2+1) + x_2(x_1+1)}{(x_1+1)(x_2+1)} = \frac{x_1 x_2 + x_1 + x_2 x_1 + x_2}{(x_1+1)(x_2+1)}
   \]
   Simplify the numerator:
   \[
   = \frac{x_1 x_2 + x_1 + x_2 x_1 + x_2}{(x_1+1)(x_2+1)} = \frac{2x_1 x_2 + x_1 + x_2}{(x_1+1)(x_2+1)}
   \]

4. Using Vieta's formulas, substitute \(x_1 + x_2 = m+3\) and \(x_1 x_2 = m+2\):
   \[
   \frac{2(m+2) + (m+3)}{(m+2) + (m+3) + 1} = \frac{3m + 7}{2m + 6}
   \]

5. Set the equation equal to \(\frac{13}{10}\) and solve for \(m\):
   \[
   \frac{3m + 7}{2m + 6} = \frac{13}{10}
   \]
   Cross-multiply to solve for \(m\):
   \[
   10(3m + 7) = 13(2m + 6)
   \]
   \[
   30m + 70 = 26m + 78
   \]
   \[
   30m - 26m = 78 - 70
   \]
   \[
   4m = 8
   \]
   \[
   m = 2
   \]

6. Verify the solution by substituting \(m = 2\) back into the original quadratic equation:
   \[
   x^2 - 5x + 4 = 0
   \]
   Factor the quadratic equation:
   \[
   (x-4)(x-1) = 0
   \]
   The roots are \(x_1 = 4\) and \(x_2 = 1\).

7. Check the condition \(\frac{x_1}{x_1+1} + \frac{x_2}{x_2+1} = \frac{13}{10}\):
   \[
   \frac{4}{4+1} + \frac{1}{1+1} = \frac{4}{5} + \frac{1}{2} = \frac{8}{10} + \frac{5}{10} = \frac{13}{10}
   \]
   The condition is satisfied.

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