In a $\triangle A B C, \frac{2\left(r_1+r_3\right)}{a c(1+\cos B)}=$

In $\triangle A B C$, if $a=8, b=10, c=12$, then $\frac{r}{R}=$

In $\triangle A B C$, if $a=13, b=8, c=7$, then $\cos (B+C)=$

In a $\triangle A B C$, if $\left(r_1-r_3\right)\left(r_1-r_2\right)-2 r_2 r_3=0$, then $a^2-b^2=$