The key to this question is to realize that we can express the value of $\frac{x_1}{x_1+1}+\frac{x_2}{x_2+1}$ in terms of $m.$ First off, we can combine the two fractions to obtain $\frac{x_1(x_2+1)+x_2(x_1+1)}{(x_2+1)(x_1+1)}=\frac{2x_1x_2+(x_1+x_2)}{(x_1x_2)+(x_1+x_2)+1}.$ We can now use Vieta's formulas to find that the previous equation is equal to $\frac{2(m+2)+(m+3)}{(m+2)+(m+3)+1}=\frac{3m+7}{2m+6}.$ Thus, we need to find all $m$ such that $$\frac{3m+7}{2m+6}=\frac{13}{10} \leftrightarrow 10(3m+7)=13(2m+6) \Leftrightarrow 30m+70=26m+78 \Leftrightarrow 4m=8 \Leftrightarrow m=2.$$After plugging this into the original equation and Checking$x^2-5x+4$ can be factored as $(x-4)(x-1),$ meaning that it has roots $4$ and $1.$ Now, we can set $x_1=4$ and $x_1=1$ and plug it into the equation to obtain $\frac{1}{2}+\frac{4}{5}=\frac{(5)(1)+(4)(2)}{10}=\frac{13}{10}.$ This means that it works and $m=2$ is indeeed a solution., we conclude that the answer is $\boxed{m=2}$.