\(3x^2-3x\left(x-2\right)=36\\ \Rightarrow3x\left(x-x+2\right)=36\\ \Rightarrow6x=36\\ \Rightarrow x=6\)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\\ \Rightarrow3x^3-4x^2+2x-1+\left(4x^2-3x^3\right)=\dfrac{5}{2}\\ \Rightarrow2x-1=\dfrac{5}{2}\\ \Rightarrow x=\dfrac{7}{4}\)