With usual notation, in a triangle ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then the value of $\cos B$ is equal to

In a triangle $A B C$, with usual notations, the sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are such that they are roots of the equation $x^3-11 x^2+38 x-40=0$ then $\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=$

In a triangle $A B C$ with usual notations if $\angle A=30^{\circ}$, then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$