\(tanx=\dfrac{1}{2}\Leftrightarrow\dfrac{sinx}{cosx}=\dfrac{1}{2}\Leftrightarrow cosx=2sinx\)
\(1+tan^2x=\dfrac{1}{cos^2x}\) \(\Leftrightarrow cos^2x=\dfrac{4}{5}\)
=> \(sin2x=2sinx.cosx=cos^2x\)
\(A=\dfrac{2sin2x}{2-3cos2x}=\dfrac{2cos^2x}{2-3\left(cos^2x-1\right)}=\dfrac{8}{13}\)