\(1,\\ a,=x\left(x-y\right)+3\left(x-y\right)=\left(x+3\right)\left(x-y\right)\\ b,=\left(x-3y\right)\left(x+3y\right)\\ c,=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ 2,\\ a,=\left[\left(2x+1\right)-\left(2x-1\right)\right]^2=2^2=4\\ 3,\\ a,x\left(x+1\right)-x\left(x-3\right)=0\\ \Leftrightarrow x\left(x+1-x+3\right)=0\\ \Leftrightarrow4x=0\Leftrightarrow x=0\\ b,x^2-6x+8=0\\ \Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,2x^2+2x+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(x^2+x+\dfrac{1}{4}\right)=0\\ \Leftrightarrow2\left(x+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
