We have

${\left(a+bx\right)}^{-2}=a^{-2}{\left[1+\frac{bx}{a}\right]}^{-2}$

                 $=a^{-2}\left(1-\frac{2bx}{a}+\frac{3b^2{x}^2}{a^2}-....\right)$

                  $=\frac{1}{a^2}-\frac{2bx}{a^3}+\frac{3b^2{x}^2}{a^4}-\dots ....\left(i\right)$

We have

${\left(a+bx\right)}^{-2}=\frac{1}{4}-3x+... ....\left(ii\right)$

Comparing $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{1}{a^2}=\frac{1}{4}\Rightarrow a=2$

And,

$\frac{2b}{a^3}=3$

$\Rightarrow 2b=3{\left(2\right)}^3\Rightarrow b=12$