1. The function $y=\log _{a}\left(x^{2}-a x+2\right)$ is always positive on $[2,+\infty)$. Then the range of the real number $a$ is $\qquad$
ii. $1.11$. Therefore, $x^{2}-a x+2$ is always greater than 1 on $[2,+\infty)$.

Secondly, from $2^{2}-2 a+2>1$, we get $a1 \text {. }$
$$

Thus, the range of values for $a$ is $1<a<\frac{5}{2}$.