2. Determine all values of the real parameter $m$ for which the equation

$$
(m+1) x^{2}-3 m x+4 m=0
$$

has two distinct real solutions that are both greater than -1.
2. Since the solutions must be real and distinct, the discriminant must be positive, i.e., $00$ and $\left(x_{1}+1\right)\left(x_{2}+1\right)>0$. From this, we have $0<x_{1}+x_{2}+2=$ $\frac{3 m}{m+1}+2=\frac{5 m+2}{m+1}$ and $0<x_{1} x_{2}+x_{1}+x_{2}+1=\frac{4 m}{m+1}+\frac{3 m}{m+1}+1=\frac{8 m+1}{m+1}$. The first inequality reduces to $m \in(-\infty,-1) \cup\left(-\frac{2}{5}, \infty\right)$, and the second to $m \in$ $(-\infty,-1) \cup\left(-\frac{1}{8}, \infty\right)$. Therefore, taking the intersection of all three obtained sets for $m$, we get that the final solution to the problem is: $m \in\left(-\frac{16}{7},-1\right) \cup\left(-\frac{1}{8}, 0\right)$.