\(a,Thaym=3.vào.\left(1\right),ta.được:x^2+5x+4=0\\ \Leftrightarrow x^2+x+4x+4=0\\ \Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;-4\right\}\\ b,\Delta=\left(m+2\right)^2-4.1.\left(m+1\right)=m^2+4m+4-4m-4=m^2\ge0\forall m\in R\\ \)