1. If for all $x$ such that $|x| \leqslant 1$, $t+1>(t^2-4)x$ always holds, then the range of values for $t$ is $\qquad$
1. $\left(\frac{\sqrt{13}-1}{2}, \frac{\sqrt{21}+1}{2}\right)$ Detailed explanation: (1) If $t^{2}-4>0$, i.e., $t>2$, then from $\frac{t+1}{t^{2}-4}>x(|x| \leqslant 1)$ always holds, we get $\frac{t+1}{t^{2}-4}>1, t+1>t^{2}-4, t^{2}-t-5-t^{2}+4 ; t^{2}+t-3>0$, solving this yields $t>\frac{-1+\sqrt{13}}{2}$, thus $\frac{-1+\sqrt{13}}{2}<t<2$. In summary, the range of values for $t$ is: $\frac{\sqrt{13}-1}{2}<t<\frac{\sqrt{21}+1}{2}$.