Triangle ABC

By sine law, sinA=a/2r, sinB=b/2r, sinC=c/2r. Thus we are required to find, sina/(sinB + sinC) = a/(b+c)

Now, subtracting the first given equation (a-4b+4c=0) from the second given equation (a+2b-3c=0), we get b=7/6c

Substituting this value of b in either of the equations, we get a=2/3c.

Substituting the values of a and b in terms of c in the required, we get a/(b+c)= 4/13

$a-4b+4c=0 \\a+2b-3c=0$

Subtract both equations and you will get

$7c=6b$ $......(k)$

Multiply 2nd eqn. by $2$ and add both of them

You will get $3a=2c$ $........(l)$

From $(k)$ and $(l)$ you can get $7a=4b$

Now by sine rule

$\dfrac{sin\angle A}{a}=\dfrac{sin\angle B}{b}=\dfrac{sin\angle C}{c}$

we get $sin\angle A=\dfrac{4 sin\angle B}{7}$

$sin\angle C=\dfrac{6 sin \angle B}{7}$

Now plugging all these values in question asked

$\dfrac{\dfrac{4 sin\angle B}{7}}{\dfrac{6 sin\angle B}{7}+sin\angle B}$ = $\dfrac{4}{13}$

Which gives us

$\dfrac {sin \angle A}{sin \angle B+sin \angle C}=\dfrac{4}{13}$

Here $m=4$ and $n=13$

$\therefore m+n \Rightarrow \huge\boxed{17}$