14. (25 points) If for any non-negative integer $n$, we have $\cos 2^{n} \alpha<-\frac{1}{3}$, find the real number $\alpha$.

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14. Prove by induction that for any $n \in \mathbf{N}$,
$$
\left|\cos 2^{n} \alpha+\frac{1}{2}\right| \geqslant\left(\frac{5}{3}\right)^{n}\left|\cos \alpha+\frac{1}{2}\right| \text {. }
$$

When $n=0$, equation (1) holds.
Assume that for $n-1$ equation (1) holds.
Then $\left|\cos 2^{n} \alpha+\frac{1}{2}\right|=2\left|\cos ^{2} 2^{n-1} \alpha-\frac{1}{4}\right|$
$$
=2\left|\cos 2^{n-1} \alpha-\frac{1}{2}\right|\left|\cos 2^{n-1} \alpha+\frac{1}{2}\right|
$$
$\geqslant 2\left(\frac{1}{3}+\frac{1}{2}\right)\left(\frac{5}{3}\right)^{n-1}\left|\cos \alpha+\frac{1}{2}\right|$
$=\left(\frac{5}{3}\right)^{n}\left|\cos \alpha+\frac{1}{2}\right|$.
By $-1 \leqslant \cos 2^{n} \alpha<-\frac{1}{3}$, then
$\left|\cos 2^{n} \alpha+\frac{1}{2}\right| \leqslant 1$.
Thus $0 \leqslant\left|\cos \alpha+\frac{1}{2}\right| \leqslant\left(\frac{3}{5}\right)^{n}(n \in \mathbf{N})$.
Since $\lim _{n \rightarrow \infty}\left(\frac{3}{5}\right)^{n}=0$, taking the limit on both sides, we must have $\cos \alpha=-\frac{1}{2}$.
Therefore, $\alpha=2 k \pi \pm \frac{2 \pi}{3}(k \in \mathbf{Z})$.