Note $2\sin |x|\cos |x|=\sin (2|x|)$. For continuity at $0$, the left-hand and right-hand limits must equal $f(0)=b$.

Right-hand limit $(x\to 0^{+})$: $|x|=x$, so

${\lim }_{x\to 0^{+}}\frac{ax+x^2-\sin 2x}{x}={\lim }_{x\to 0^{+}}\left(a+x-\frac{\sin 2x}{x}\right)=a+0-2=a-2$.

Left-hand limit $(x\to 0^{-})$: $|x|=-x$, so

${\lim }_{x\to 0^{-}}\frac{-ax+x^2-\sin (-2x)}{x}={\lim }_{x\to 0^{-}}\left(-a+x+\frac{\sin 2x}{x}\right)=-a+0+2=2-a$.

Setting LHL = RHL: $a-2=2-a\Rightarrow a=2$, and so $b=0$.

$\therefore a+b=2$.