Five. (20 points) Given the functions
$$
\begin{array}{l}
f(x)=x \ln x, \\
g(x)=-x^{2}+a x-3(a \in \mathbf{R}) .
\end{array}
$$
(1) If for any $x \in(0,+\infty)$, the inequality $f(x) \geqslant \frac{1}{2} g(x)$ always holds, find the range of $a$;
(2) Prove that for any $x \in(0,+\infty)$, we have
$$
\ln x>\frac{1}{\mathrm{e}^{x}}-\frac{2}{\mathrm{e} x} .
$$
(1) When $x>0$,
$$
\begin{array}{l}
f(x) \geqslant \frac{1}{2} g(x) \\
\Leftrightarrow \ln x \geqslant \frac{1}{2}\left(-x^{2}+a x-3\right) \\
\Leftrightarrow a \leqslant 2 \ln x+x+\frac{3}{x} . \\
\text { Let } h(x)=2 \ln x+x+\frac{3}{x}(x>0) .
\end{array}
$$

Then $a \leqslant h(x)_{\text {min }}$.
By $h^{\prime}(x)=\frac{(x+3)(x-1)}{x^{2}}$, we know that the function $h(x)$ is monotonically decreasing on the interval $(0,1)$ and monotonically increasing on the interval $(1,+\infty)$.
Therefore, $h(x)_{\min }=h(1)=4$.
Hence, the range of $a$ is $(-\infty, 4]$.
(2) To prove $\ln x>\frac{1}{\mathrm{e}^{x}}-\frac{2}{\mathrm{ex}}$, we only need to prove
$$
f(x)=x \ln x>\frac{x}{\mathrm{e}^{x}}-\frac{2}{\mathrm{e}} \text {. }
$$

By $f^{\prime}(x)=\ln x+1$, we know that $f(x)$ is monotonically decreasing on the interval $\left(0, \frac{1}{\mathrm{e}}\right)$ and monotonically increasing on the interval $\left(\frac{1}{\mathrm{e}},+\infty\right)$.
Thus, when $x>0$,
$$
f(x) \geqslant f\left(\frac{1}{\mathrm{e}}\right)=-\frac{1}{\mathrm{e}} \text {. }
$$

Let $\varphi(x)=\frac{x}{\mathrm{e}^{x}}-\frac{2}{\mathrm{e}}(x>0)$. Then
$$
\varphi^{\prime}(x)=\frac{1-x}{\mathrm{e}^{x}} .
$$

Therefore, $\varphi(x)$ is monotonically increasing on the interval $(0,1)$ and monotonically decreasing on the interval $(1,+\infty)$.
Thus, $\varphi(x) \leqslant \varphi(1)=-\frac{1}{\mathrm{e}}$.
Clearly, the equalities in (1) and (2) cannot hold simultaneously.
Hence, when $x>0$, $f(x)>\varphi(x)$, i.e.,
$$
\ln x>\frac{1}{\mathrm{e}^{x}}-\frac{2}{\mathrm{e} x} \text {. }
$$