$-3<\frac{x^2-\lambda x-2}{x^2+x+1}<2$
$\Rightarrow -3x^2-3x-3<x^2-\lambda x-2<2x^2+2x+2$ $\left(\mathrm{since} x^2+x+1>0, \forall x\in R\right)$
$\Rightarrow 4x^2+x\left(3-\lambda \right)+1>0,x^2+x\left(2+\lambda \right)+4>0$
$\left(i\right) 4x^2-x\left(\lambda -3\right)+1>0$
$\Rightarrow D<0\Rightarrow {\left(\lambda -3\right)}^2-4\times 4\times 1<0$
$\Rightarrow \left(\lambda -3+4\right)\left(\lambda -3-4\right)<0$
$\Rightarrow \left(\lambda +1\right)\left(\lambda -7\right)<0\mathrm{}\mathrm{}\Rightarrow \lambda \in \left(-\mathrm{1,7}\right)$
$\left(ii\right) x^2+x\left(\lambda +2\right)+4>0$
$\Rightarrow D<0\Rightarrow {\left(\lambda +2\right)}^2-4\times 4<0$
$\begin{array}{l}\left(\lambda +2-4\right)\left(\lambda +2+4\right)<0\\ \left(\lambda -2\right)\left(\lambda +6\right)<0\Rightarrow \lambda \in \left(-\mathrm{6,2}\right)\end{array}$
Taking the intersection of the solutions of $\left(i\right)$ and $\left(ii\right)$, we get, 

 $\lambda \in \left(-\mathrm{1,2}\right)$