12. Given vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ satisfy $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$, and the angle between them is $60^{\circ}$. Then the range of real number $\lambda$ that makes the angle between vectors $\boldsymbol{a}+\lambda \boldsymbol{b}$ and $\lambda \boldsymbol{a}-2 \boldsymbol{b}$ obtuse is $\qquad$
12. $(-1-\sqrt{3},-1+\sqrt{3})$.

From the problem, we get $(a+\lambda b) \cdot(\lambda a-2 b)=\lambda^{2}+2 \lambda-2<0$.
Solving this, we get $-1-\sqrt{3}<\lambda<-1+\sqrt{3}$.