2. First method: Since $x_{1}^{3}-2015 x_{1}+2016=0$ and $x_{2}^{3}-2015 x_{2}+2016=0$, then $x_{1}^{3}-x_{2}^{3}=2015 \cdot\left(x_{1}-x_{2}\right)$. Therefore, $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=2015$.

Second method: By Vieta's theorem (but then it needs to be justified that there are three distinct roots): $x_{1}+x_{2}+x_{3}=0, x_{1} x_{2} x_{3}=-2016$. Therefore,

$$
\begin{aligned}
x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=\left(-x_{3}\right)^{2}+\frac{2016}{x_{3}} & \\
=\frac{x_{3}^{3}+2016}{x_{3}} & =\frac{2015 x_{3}}{x_{3}}=2015
\end{aligned}
$$
Answer: 2015.

Answer to option: 4-2: 2016.