(3) $(1,4]$ Hint: Let $g(x)=4 x+\frac{a}{x}$. From $f(x)=\log _{a}\left(4 x+\frac{a}{x}\right)$ being an increasing function on the interval $[1,2]$, we know that when $0<a<1$, $g(x)$ is a decreasing function on the interval $[1,2]$, and $g(x)>0$, thus $g'(x) \leqslant 0$, therefore $4-\frac{a}{x^2} \leqslant 0$, hence $\frac{\sqrt{a}}{2} \geqslant 2, a$ does not exist.

When $a>1$, $g(x)$ is an increasing function on the interval $[1,2]$, and $g(x)>0$, thus $\frac{\sqrt{a}}{2} \leqslant 1, a \leqslant 4$.
Therefore, the range of values for $a$ is $(1,4]$.
(3) If the function $f(x)=\log _{a}\left(4 x+\frac{a}{x}\right)$ is increasing on the interval $[1$,
2], then the range of values for $a$ is $\qquad$