FINAL ROUND

Answer: 4. Note that 1 is the root of the equation $$x^3-3x^2+(a+2)x-a=x^3-x^2-2x^2+2x+ax-a=(x-1)(x^2-2x+a),(1)$$ Therefore, since $$x^3-3x^2+(a+2)x-a=x^3-x^2-2x^2+2x+ax-a=(x-1)(x^2-2x+a),$$ we obtain the equivalent equation $(x-1)(x^2-2x+a)=0$. So, two other roots of equation (1) are the roots of the equation $$x^2-2x+a=0,(2)$$ their sum is equal to 2, so one of them is less than 1 and another is greater than 1. Hence $x_2=1$, and $x_1$, $x_3$ are the roots of (2). Therefore, $$\begin{gathered} 4x_1-x_1^2+x_3^2=(x_3+x_1)(x_3-x_1)+4x_1= \\ =2(x_3-x_1)+4x_1=2(x_3+x_1)=2\cdot 2=4. \end{gathered}$$