3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ is a prime number, then the range of values for the real number $x$ is
3. $-\frac{3}{2} \leqslant x<-\frac{1}{2}$ or $\frac{3}{2} \leqslant x<\frac{5}{2}$.

Since $\left[x-\frac{1}{2}\right]$ and $\left[x+\frac{1}{2}\right]$ are both integers, to make $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ a prime number, one of $\left[x-\frac{1}{2}\right]$ or $\left[x+\frac{1}{2}\right]$ must be 1 or -1.
When $\left[x-\frac{1}{2}\right]=1$,
$$
1 \leqslant x-\frac{1}{2}<2, \frac{3}{2} \leqslant x<\frac{5}{2},
$$

At this time, $\left[x+\frac{1}{2}\right]=2$, which satisfies the condition;
When $\left[x+\frac{1}{2}\right]=-1$,
$-1 \leqslant x+\frac{1}{2}<0,-\frac{3}{2} \leqslant x<-\frac{1}{2}$,

At this time, $\left[x-\frac{1}{2}\right]=-2$, which satisfies the condition;
When $\left[x+\frac{1}{2}\right]=1$ or $\left[x-\frac{1}{2}\right]=-1$, it is easy to see that $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ is not a prime number.