Four. (Full marks 20 points) Let $t_{1}, t_{2}$ be the two distinct real roots of the equation $t^{2}-$ $(5 a-2) t-3 a^{2}-7 a+1=0$. Find the value of the real number $a$ such that for any non-zero real number $m$, the function $f(x)=\cos (m \pi x) \cos \left[\left(t_{1}^{3}+t_{2}^{3}\right) \cdot \pi x\right]$ is a periodic function.
Let $T$ be the smallest positive period of the function $f(x)$, then we have $f(T) = f(0)$, i.e.,
$$
\cos (m \pi T) \cos \left(\left(t_{1}^{3} + t_{2}^{3}\right) \pi T\right) = 1.
$$
By the boundedness of the cosine function, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
|\cos (m \pi T)| = 1, \\
\left|\cos \left[\left(t_{1}^{3} + t_{2}^{3}\right) \pi T\right]\right| = 1.
\end{array}\right. \\
\therefore \left\{\begin{array}{l}
m \pi T = k_{1} \pi, \\
\left(t_{1}^{3} + t_{2}^{3}\right) \pi T = k_{2} \pi.
\end{array} \quad (k_{1}, k_{2} \in \mathbb{Z}, k_{1} \neq 0)\right. \\
\therefore \frac{t_{1}^{3} + t_{2}^{3}}{m} = \frac{k_{2}}{k_{1}}.
\end{array}
$$

That is, for any non-zero real number $m$, $\frac{t_{1}^{3} + t_{2}^{3}}{m}$ is a rational number.
Therefore, we must have $t_{1}^{3} + t_{2}^{3} = 0$, i.e.,
$$
\begin{array}{l}
\left(t_{1} + t_{2}\right)\left(t_{1}^{2} - t_{1} t_{2} + t_{2}^{2}\right) = 0. \\
\because t_{1}, t_{2} \in \mathbb{R}, \text{ and } t_{1} \neq t_{2}, \\
\therefore t_{1}^{2} - t_{1} t_{2} + t_{2}^{2} \neq 0.
\end{array}
$$

Hence, $t_{1} + t_{2} = 0$.
By Vieta's formulas, we get $(5a - 2)^{2} = 0$. Thus, $a = \frac{2}{5}$.