5. (11th "Hope Cup" Invitational Competition Question) When $0<\alpha<\frac{\pi}{2}$, the range of $(\sin \alpha+\tan \alpha)(\cos \alpha+\cot \alpha)$ is $\qquad$ .
5. Fill $\left(2, \frac{\sqrt{3}}{2}+\sqrt{2}\right]$. Reason: It is easy to know that $(\sin \alpha+\tan \alpha)(\cos \alpha+\cot \alpha)=\sin \alpha \cdot \cos \alpha+\sin \alpha+\cos \alpha+1=\frac{1}{2}(\sin \alpha+$ $\cos \alpha)^{2}+(\sin \alpha+\cos \alpha)+\frac{1}{2}$.

Let $\sin \alpha+\cos \alpha=t$, then the above expression becomes $\frac{1}{2} t^{2}+t+\frac{1}{2}=\frac{1}{2}(t+1)^{2}$. Since $0<\alpha<\frac{\pi}{2}$, then $t=\sqrt{2} \sin \left(\frac{\pi}{4}+\alpha\right) \in(1$, $\sqrt{2}]$, hence $2<\frac{1}{2}(t+1)^{2} \leqslant \frac{3}{2}+\sqrt{2}$.