Since, $f\left(x\right)$ is continuous at $x=0$ so, at $x=0$ both left hand and right hand limits must exist and both must be equal to $3.$
Now,
$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{a\left(1-x\sin x\right)+b\cos x+5}{x^2}=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{\lim }\frac{a-ax\left(x-\frac{x^3}{3!}+....\right)+b\left(1-\frac{x^2}{2!}+....\right)+5}{x^2}=f\left(0\right)$
$\Rightarrow \underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(a+b+5\right)+\left(-a-\frac{b}{2}\right)x^2+...}{x^2}=3$
If $\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=3$ exists, then $a+b+5=0$
and $-a-\frac{b}{2}=3$

$\Rightarrow a=-1$ and $b=-4$