4. If the system of inequalities about $x$ $\left\{\begin{array}{l}x^{3}+3 x^{2}-x-3>0, \\ x^{2}-2 a x-1 \leqslant 0\end{array},(a>0)\right.$ has exactly one integer solution, then the range of values for $a$ is $\qquad$
Solution: $\left[\frac{3}{4}, \frac{4}{3}\right)$.
From $x^{3}+3 x^{2}-x-3>0$, we get $-31$.
Therefore, the only possible integer solutions to the inequality system are -2 or 2.
Since the axis of symmetry of the function $f(x)=x^{2}-2 a x-1$ is $x=a>0$, the integer solution to the inequality system can only be 2. Therefore, we have
$$
\left\{\begin{array}{l}
f(-2)=3+4 a>0, \\
f(2)=3-4 a \leqslant 0, \\
f(3)=8-6 a>0 .
\end{array} \text { Solving, we get } \frac{3}{4} \leqslant a<\frac{4}{3}\right. \text { . }
$$

Thus, the range of values for $a$ is $\left[\frac{3}{4}, \frac{4}{3}\right)$.