a) \(f\left(-2\right)=1\Leftrightarrow2+m-2=1\Rightarrow m=1\)
b) \(\left\{{}\begin{matrix}y=-x+m-2\left(d\right)\\y=x^2\left(p\right)\end{matrix}\right.\)
\(\Rightarrow-x+m-2=x^2\Leftrightarrow x^2+x-m+2=0\)
Theo định lí Viete ta có : \(\left\{{}\begin{matrix}x_1x_2=\dfrac{m+2}{1}=m+2\\x_1+x_2=\dfrac{-1}{1}=-1\end{matrix}\right.\)
Ta có \(x_1x_2-4x_1-4x_2=x_1x_2-4\cdot\left(x_1+x_2\right)=4\)
\(\Leftrightarrow m+2-4\cdot\left(-1\right)=4\Rightarrow m=-2\)
Vậy m = -2 thì \(x_1x_2-4x_1-4x_2=4\)