22 (12 points) Let $f(x)=\frac{a}{\sim}+x \ln x, g(x)=x^{3}-x^{2}-3$.
(1) When $a=2$, find the equation of the tangent line to the curve $y=f(x)$ at $x=1$;
(2) If there exist $x_{1}, x_{2} \in[0,2]$ such that $g\left(x_{1}\right)-g\left(x_{2}\right) \geqslant M$ holds, find the largest integer $M$ that satisfies the above condition;
(3) If for any $s, t \in\left[\frac{1}{2}, 2\right]$ we have $f(s) \geqslant g(t)$, find the range of real numbers $a$.
22 (1) When $a=2$, $f(x)=\frac{2}{x}+x \ln x, f^{\prime}(x)=-\frac{2}{x^{2}}+\ln x+1, f(1)=2, f^{\prime}(1)=-1$. Therefore,
$$
y-2=-(x-1).
$$

So the equation of the tangent line to the curve $y=f(x)$ at $x=1$ is $x+y-3=0$.
(2) There exist $x_{1}, x_{2} \in[0,2]$, such that $g\left(x_{1}\right)-g\left(x_{2}\right) \geqslant M$ holds, which is equivalent to: $\left[g\left(x_{1}\right)-g\left(x_{2}\right)\right]_{\max } \geqslant M$.
$$
g(x)=x^{3}-x^{2}-3, g^{\prime}(x)=3 x^{2}-2 x=3 x\left(x-\frac{2}{3}\right),
$$
\begin{tabular}{|c|c|c|c|c|c|}
\hline$x$ & 0 & $\left(0, \frac{2}{3}\right)$ & $\frac{2}{3}$ & $\left(\frac{2}{3}, 2\right)$ & 2 \\
\hline$g^{\prime}(x)$ & & - & 0 & + & \\
\hline$g(x)$ & -3 & decreasing & minimum value $-\frac{85}{27}$ & increasing & 1 \\
\hline
\end{tabular}

From the table above, we know:
$$
\begin{array}{c}
g(x)_{\min }=g\left(\frac{2}{3}\right)=-\frac{85}{27}, \\
g(x)_{\max }=g(2)=1, \\
{\left[g\left(x_{1}\right)-g\left(x_{2}\right)\right]_{\max }=g(x)_{\max }-g(x)_{\min }=\frac{112}{27},}
\end{array}
$$

Therefore, the largest integer $M$ that satisfies the condition is $4$.
(3) For any $s, t \in\left[\frac{1}{2}, 2\right]$, $f(s) \geqslant g(t)$ holds. This is equivalent to: on the interval $\left[\frac{1}{2}, 2\right]$, the minimum value of the function $f(x)$ is not less than the maximum value of $g(x)$. From (2), we know that on the interval $\left[\frac{1}{2}, 2\right]$, the maximum value of $g(x)$ is $g(2)=1$.
So $f(x)_{\min } \geqslant 1$. Since $f(1)=a$, we have $a \geqslant 1$.
Next, we prove that when $a \geqslant 1$, on the interval $\left[\frac{1}{2}, 2\right]$, the function $f(x) \geqslant 1$ holds.
When $a \geqslant 1$ and $x \in\left[\frac{1}{2}, 2\right]$,
$$
f(x)=\frac{a}{x}+x \ln x \geqslant \frac{1}{x}+x \ln x.
$$

Let $h(x)=\frac{1}{x}+x \ln x, h^{\prime}(x)=-\frac{1}{x^{2}}+\ln x+1, h^{\prime}(1)=0$.
When $x \in\left[\frac{1}{2}, 1\right]$, $h^{\prime}(x)=-\frac{1}{x^{2}}+\ln x+1 \geq 0$.
So the function $h(x)=\frac{1}{x}+x \ln x$ is decreasing on the interval $\left[\frac{1}{2}, 1\right)$ and increasing on the interval $(1,2]$, $h(x)_{\min }=h(1)=1$, i.e., $h(x) \geqslant 1$, so when $a \geqslant 1$ and $x \in\left[\frac{1}{2}, 2\right]$, $f(x) \geqslant 1$ holds, i.e., for any $s, t \in\left[\frac{1}{2}, 2\right]$, $f(s) \geqslant g(t)$.