(512 (1) Given the function $y=\left(\frac{1}{3}\right)^{x}$ with the domain of the independent variable $x$ being $\{x \mid x \geqslant-1\}$, find the corresponding range of function values;
(2) Find the range of the function $y=2^{x^{2}+1 x}$.

Solution (1) Since $y=\left(\frac{1}{3}\right)^{x}$ is a decreasing function on $(-\infty,+\infty)$, and $x \geqslant-1$, we get
$$
\left(\frac{1}{3}\right)^{x} \leqslant\left(\frac{1}{3}\right)^{-1},
$$

i.e., $0<y \leqslant 3$, the range of the function values is $\{y \mid 0<y \leqslant 3\}$.
(2) The function $y=2^{\prime}$ is an increasing function on $(-\infty,+\infty)$. Since
$$
x^{2}+4 x=(x+2)^{2}-4 \geqslant-4,
$$

we get $2^{x^{2}+4 x} \geqslant 2^{-4}=\frac{1}{16}$, i.e., $y \geqslant \frac{1}{16}$, the range is $\left[\frac{1}{16},+\infty\right)$.

Solution Guidance When studying the properties of composite functions, it is always converted into several basic functions, i.e., transforming complex problems into simple problems or problems that have already been solved. The following problem is also the same.