14. Given the function $f(x)=4 x+a x^{2}-\frac{2}{3} x^{3}$, where $a \in[-1,1]$ is a constant. Let the two non-zero real roots of the equation $f(x)=2 x+\frac{1}{3} x^{3}$ with respect to $x$ be $x_{1}, x_{2}$. Is there a real number $m$ such that the inequality $m^{2}+t m+1 \geqslant \left|x_{1}-x_{2}\right|$ holds for any $a \in[-1,1]$ and $t \in[-1,1]$? If it exists, find the range of $m$; if not, explain the reason.
14. $4 x+a x^{2}-\frac{2}{3} x^{3}=2 x+\frac{1}{3} x^{3}$ yields: $x=0$ or $x^{2}-a x-2=0$. Since $\Delta=a^{2}+8>0$, $x_{1}, x_{2}$ are the two non-zero real roots of the equation $x^{2}-a x-2=0$, with $x_{1}+x_{2}=a, x_{1} x_{2}=-2$. Therefore, $\left|x_{1}-x_{2}\right|=$ $\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{a^{2}+8}$. Since $-1 \leqslant a \leqslant 1$, we have $\left|x_{1}-x_{2}\right|=\sqrt{a^{2}+8} \leqslant 3$.

To ensure the inequality $m^{2}+t m+1 \geqslant\left|x_{1}-x_{2}\right|$ holds for $a \in[-1,1]$ and $t \in[-1,1]$, it is necessary and sufficient that $m^{2}+t m+1 \geqslant 3$ for any $t \in[-1,1]$, i.e., $m^{2}+t m-2 \geqslant 0$.

Let $g(t)=m^{2}+t m-2=m t+\left(m^{2}-2\right)$. Then $g(-1)=m^{2}-m-2 \geqslant 0$ and $g(1)=m^{2}+m-2 \geqslant 0$. Solving these, we get: $m \geqslant 2$ or $m \leqslant-2$. Therefore, there exists $m$ such that $m^{2}+t m+1 \geqslant\left|x_{1}-x_{2}\right|$ holds for any $a \in[-1,1]$ and $t \in[-1,1]$, with the range of $m$ being $\{m \mid m \geqslant 2$ or $m \leqslant-2\}$.