Domain: $\mathbb{R}$; Range: $(-\infty, 0]$.

The modulus function $|x|$ is defined for all real numbers, and $|x| \geq 0$. Therefore, $f(x) = -|x| \leq 0$, so the range consists of all non-positive real numbers.

Domain: $[-3, 3]$; Range: $[0, 3]$.

For $f(x) = \sqrt{9-x^2}$ to be real, we require $9-x^2 \geq 0$, i.e., $x^2 \leq 9$, which gives $-3 \leq x \leq 3$. The maximum value of $f(x)$ is $3$ (at $x=0$) and the minimum is $0$ (at $x=\pm 3$). Since the function is continuous, it takes all values between $0$ and $3$.