$\begin{aligned} & \text {} \because x^3+2 x+5=0 \\ & \alpha+\beta+\gamma=0 \\ & \alpha \beta+\beta \gamma+\gamma \alpha=2 \\ & \alpha \beta \gamma=-5 \\ & \sum \frac{\beta+\gamma}{\alpha^2}=\frac{\beta+\gamma}{\alpha^2}+\frac{\alpha+\gamma}{\beta^2}+\frac{\alpha+\beta}{\gamma^2} \\ & =\frac{\beta+\gamma+\alpha}{\alpha^2}-\frac{1}{\alpha}+\frac{\alpha+\beta+\gamma}{\beta^2}-\frac{1}{\beta}+\frac{\alpha+\beta+\gamma}{\gamma^2}-\frac{1}{\gamma} \\ & =\alpha+\beta+\gamma\left[\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right]-\left[\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right] \\ & =(\alpha+\beta+\gamma)\left[\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{\alpha^2 \beta^2 \gamma^2}\right]-\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma} \\ & =0 \times\left[\frac{\beta^2 \gamma^2+\alpha^2 \gamma^2+\alpha^2 \beta^2}{\alpha^2 \beta^2 \gamma^2}\right]-\frac{2}{-5}=\frac{2}{5}\end{aligned}$