5. Let vectors $a$ and $b$ satisfy $|a|=1,|b|=2$, and the angle between $a$ and $b$ is $60^{\circ}$. If the angle between the vectors $7 a+2 t b$ and $t a+b$ is obtuse, then the range of the real number $t$ is $\qquad$
5. $-7<t<-\frac{1}{2}$, and $t \neq-\frac{\sqrt{14}}{2}$.

From $(7 a+2 t b) \cdot(t a+b)<0$, we get $-7<t<-\frac{1}{2}$.
Moreover, the angle between $7 a+2 t b$ and $t a+b$ must not be a straight angle, hence, $t \neq-\frac{\sqrt{14}}{2}$.