\textbf{法一：} 由 $\tan \theta=-2$，得 $\sin^2\theta=\frac{4}{5}$，$\sin \theta\cdot\cos\theta=-\frac{2}{5}$．故 $\frac{\sin \theta(1+\sin 2\theta)}{\sin\theta+\cos\theta}=\frac{\sin \theta(\sin \theta+\cos\theta)^2}{\sin\theta+\cos\theta}=\sin^2\theta+\sin\theta\cos\theta=\frac{2}{5}$
\textbf{法二：} $\frac{\sin \theta(1+\sin 2\theta)}{\sin\theta+\cos\theta}=\frac{\sin \theta(\sin \theta+\cos\theta)^2}{\sin\theta+\cos\theta}=\frac{\sin^2\theta+\sin\theta\cos\theta}{\sin^2\theta+\cos^2\theta}=\frac{\tan^2\theta+\tan\theta}{\tan^2\theta+1}=\frac{2}{5}$．选 $C$