Answer
$(-\infty,-4)U(2,4)U(6,\infty)$. See graph.
Work Step by Step
Step 1. $\frac{x^2-8x+12}{x^2-16}\gt0 \longrightarrow \frac{(x-2)(x-6)}{(x+4)(x-4)}\gt0 $
Step 2. Identify boundary points $x=-4,2,4,6$ and intervals $(-\infty,-4),(-4,2),(2,4),(4,6),(6,\infty)$
Step 3. Choose test values $x=-5,0,3,5,7$ for each interval and evaluate the inequality to get $True,\ False,\ True,\ False,\ True$
Step 4. Thus we have the solution interval(s) $(-\infty,-4)U(2,4)U(6,\infty)$. See graph.