If \[1+a-b\ne 0,\]then L.H.S. \[\to \infty \] as \[x\to 0\] while R.H.S. =1, therefore \[1+a-b=0.\]
Now from (i), \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{x}^{2}}\left( \frac{b}{3\,\,!}-\frac{a}{2\,\,!} \right)+{{x}^{4}}\,\left( \frac{a}{4\,\,!}-\frac{b}{5\,\,!} \right)+...}{{{x}^{2}}}=1\]
\[\Rightarrow \,\,\frac{b}{3\,!}-\frac{a}{2\,\,!}=1\,\Rightarrow \,\,b-3a=6\]. Solving \[1+a-b=0\] and \[b-3a=6,\] we get \[a=-5/2,\,\,b=-3/2\].