12.49 At what points are the tangents to the curve $y=\frac{x^{3}}{3}-x^{2}-x+1$ parallel to the line $y=2 x-1$?
12.49 Let $\left(x_{0} ; y_{0}\right)$ be the point of tangency. The slope of the tangent line at this point is $k=2$. We find $y^{\prime}=x^{2}-2 x-1 ; \quad k=y^{\prime}\left(x_{0}\right)=x_{0}^{2}-2 x_{0}-1=2 . \quad$ Solving the equation $x_{0}^{2}-2 x_{0}-3=0$, we get $\left(x_{0}\right)_{1}=3,\left(x_{0}\right)_{2}=-1$,
from which $\left(y_{0}\right)_{1}=-2,\left(y_{0}\right)_{2}=\frac{2}{3}$. Therefore, the points of tangency are $(3 ;-2)$ and $\left(-1 ; \frac{2}{3}\right)$.

Answer: $(3 ;-2)$ and $\left(-1 ; \frac{2}{3}\right)$.