Answer
$1.\ \ \mathrm{Domain:}\ \ (-\infty ,-4)\cup (-4,4)\cup (4,\infty )$
$2.\ \ \mathrm{Range:}\ \ (-\infty ,-\frac{1}{8}]\cup (0,\infty )$
Work Step by Step
$1.$ Domain of the rational number must not be zero. So take the denominator and compare it to zero to find the undefined points.
$t^2-16=0$, $\ \ $ solve for $x$.
$t^2=16$
For $\ \ x^2=f(a)\ \ $ the solutions are $\ \ x=\sqrt{f(a)}\ and-\sqrt{f(a)}$
$t=\sqrt{16}\ and \ t=-\sqrt{16}$
$t=4\ and \ t=-4$
These are the undefined points.
So, the function domain is, $-4< t <4$, which, in interval notation, can be represented as, $\ \ (-\infty ,-4)\cup (-4,4)\cup (4,\infty )$
$2.$ The smallest value the denominator can take is -16. So, when the denominator is negative, the function can only take on values between $\ \ -\infty\ \ $ and $\ \ \frac{-2}{16}\ \ $, not between $\ \ -\infty\ \ $ and $\ \ 0$.
So the range of the function is, $(-\infty ,-\frac{1}{8}]\cup (0,\infty )$
