14. Given the function $f(x)=x \ln x-\frac{1}{3} a x^{3}+2 x(a>0)$.
(1) If $f(x) \leqslant 1$ always holds, find the minimum value of $a$;
(2) If $f(x)$ has a maximum value, find the range of values for $a$.
(1) When $x=1$, $-\frac{1}{3} a+2 \leqslant 1 \Rightarrow a \geqslant 3$. We now prove that $x \ln x-x^{3}+2 x \geqslant 1$ always holds.

Let $g(x)=\ln x-x^{2}+2-\frac{1}{x}$, $g^{\prime}(x)=\frac{1}{x}-2 x+\frac{1}{x^{2}}=\frac{-2 x^{3}+x+1}{x^{2}}$ $=\frac{-2(x-1)\left(2 x^{2}+x+1\right)}{x^{2}}$, then $g(x)$ is monotonically increasing on $(0,1)$ and monotonically decreasing on $(1,+\infty)$, so $g(x)_{\max }=g(1)=0$. Therefore, $g(x) \leqslant 0 \Rightarrow \ln x-x^{2}+2 \leqslant \frac{1}{x} \Rightarrow f(x) \leqslant 1$.
(2) $f^{\prime}(x)=\ln x-a x^{2}+3=x^{2}\left(\frac{\ln x+3}{x^{2}}-a\right)$. Let $h(x)=\frac{\ln x+3}{x^{2}}$, $h^{\prime}(x)=\frac{1-2(\ln x+3)}{x^{3}}=-\frac{2 \ln x+5}{x^{3}}$, then $h(x)$ is monotonically increasing on $\left(0, \mathrm{e}^{-\frac{5}{2}}\right)$ and monotonically decreasing on $\left(\mathrm{e}^{-\frac{5}{2}},+\infty\right)$, so $h(x)_{\max }=h\left(\mathrm{e}^{-\frac{5}{2}}\right)=\frac{\mathrm{e}^{5}}{2}$.
(1) When $a \geqslant \frac{\mathrm{e}^{5}}{2}$, $f(x)$ is monotonically decreasing on $(0,+\infty)$, and has no maximum value;
(2) When $0<a<\frac{\mathrm{e}^{5}}{2}$, as $x \rightarrow 0$, $h(x) \rightarrow-\infty$; as $x \rightarrow+\infty$, $h(x) \rightarrow 0$. Then $h(x)-a=0$ has two roots, denoted as $x_{1}<x_{2}$, thus $f(x)$ is monotonically decreasing on $\left(0, x_{1}\right)$, monotonically increasing on $\left(x_{1}, x_{2}\right)$, and monotonically decreasing on $\left(x_{2},+\infty\right)$, i.e., $f(x)$ has a maximum value $f\left(x_{2}\right)$.
Noting that as $x \rightarrow 0$, $f(x) \rightarrow 0$, then $f(x)$ has a maximum value $\Leftrightarrow f\left(x_{2}\right) \geqslant 0 \Leftrightarrow\left\{\begin{array}{l}\frac{\ln x_{2}+3}{x_{2}^{2}}=a, \\ \ln x_{2}-\frac{1}{3} a x_{2}^{2}+2 \geqslant 0\end{array}\right.$ $\Rightarrow \frac{2}{3} \ln x_{2}+1 \geqslant 0 \Rightarrow x_{2} \geqslant \mathrm{e}^{-\frac{3}{2}}$. Since $a=h\left(x_{2}\right)$ and $h(x)$ is monotonically decreasing on $\left(\mathrm{e}^{-\frac{5}{2}},+\infty\right)$, we get $0<a \leqslant \frac{3}{2} \mathrm{e}^{3}$. In summary, the range of $a$ is $\left(0, \frac{3}{2} \mathrm{e}^{3}\right]$.