\(a,\Leftrightarrow f\left(x\right)⋮g\left(x\right)=\left(x+2\right)^2\\ \Leftrightarrow f\left(-2\right)=-8+4a-4=0\\ \Leftrightarrow a=3\\ b,\Leftrightarrow f\left(x\right)⋮g\left(x\right)=\left(x-1\right)\left(x+1\right)\\ \Leftrightarrow f\left(1\right)=f\left(-1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}1+a+b-1=0\\1-a-b-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=0\\a+b=0\end{matrix}\right.\Leftrightarrow a,b\in R\\ \text{Vậy }f\left(x\right)⋮g\left(x\right),\forall a,b\\ c,\Leftrightarrow f\left(1\right)=f\left(-2\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2-3a+2+b=0\\-18-12a-4+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-b=4\\12a-b=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{26}{9}\\b=-\dfrac{38}{3}\end{matrix}\right.\)