What you have written as the dot product isn't the dot product.
The dot product of $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ is simply $v\cdot w=v_1w_1+v_2w_2+v_3w_3.$
In your problem, we have $w=(0,-2,-2)$. Computing $-2w$, we have $-2w=-2(0,-2,-2)=\bigl(-2\cdot0, -2\cdot(-2), -2\cdot(-2)\bigr)=(0,4,4).$ Then $(-2w)\cdot w= (0,4,4)\cdot(0,-2,-2) =0\cdot0+(4)\cdot(-2)+(4)\cdot(-2)=-16.$
Your expression at the end of your post demands that we find the norm of two vectors. The norm of $w$ is $\Vert w\Vert=\sqrt{ 0^2+(-2)^2+(-2)^2}=\sqrt {8}=2\sqrt2.$
The norm of $-2w$ can be calculated in a similar manner; but you could also compute it using the rule $\Vert \alpha v\Vert=|\alpha|\Vert v\Vert$:
$\Vert -2w\Vert=2\Vert w\Vert=2(2\sqrt2)=4\sqrt2.$
Finally, if you wish to compute $(-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert$, just substitute: $ { (-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert} ={-16\over (2\sqrt2)(4\sqrt2)}=-1. $
This should come as no suprise, since $ (-2w)\cdot w =-2(w\cdot w)=-2\Vert w\Vert^2$ and $\Vert -2w\Vert =2\Vert w\Vert$; we could have arrived at the result with no computations: $ { (-2w)\cdot w\over \Vert w\Vert \Vert 2w\Vert} ={-2\Vert w\Vert^2 \over2\Vert w\Vert\Vert w\Vert}={-2\Vert w\Vert^2 \over2\Vert w\Vert^2}=-1. $