\(1,=4x^2-1\\ 2,=\left(x-4\right)^2-9y^2=\left(x-3y-4\right)\left(x+3y-4\right)\)
1)\(\left(2x+1\right)\left(2x-1\right)=\left(2x\right)^2-1^2=4x^2-1\)
2)\(x^2-8x-9y^2+16=\left(x^2-8x+16\right)-9y^2=\left(x^2-8x+4^2\right)-\left(3y\right)^2=\left(x-4\right)^2-\left(3y\right)^2=\left[\left(x-4\right)-3y\right]\left[\left(x-4\right)+3y\right]=\left(x-4-3y\right)\left(x-4+3y\right)\)