5. Given vectors $|\vec{a}|=1,|\vec{b}|=2$, and the angle between $\vec{a}, \vec{b}$ is $120^{\circ}$. If the angle between $\vec{a}+t \vec{b}$ and $t \vec{a}+\vec{b}$ is acute, then the range of $t$ is $\qquad$.
$$
\begin{array}{l}
(\vec{a}+t \vec{b}) \cdot(t \vec{a}+\vec{b})=t \vec{a}^{2}+\left(t^{2}+1\right) \vec{a} \cdot \vec{b}+t \vec{b}^{2} \\
=t+4 t+1 \times 2 \times\left(-\frac{1}{2}\right)\left(t^{2}+1\right)=-t^{2}+5 t-1>0 \Rightarrow \frac{5-\sqrt{21}}{2}<t<\frac{5+\sqrt{21}}{2},
\end{array}
$$

Also, $(\vec{a}+t \vec{b}) \parallel (t \vec{a}+\vec{b}) \Rightarrow t^{2}=1$ is not suitable for the problem.
Therefore, the range of $t$ is $\left(\frac{5-\sqrt{21}}{2}, 1\right) \cup\left(1, \frac{5+\sqrt{21}}{2}\right)$.