3. If $x \in(-1,1)$, then
$$
f(x)=x^{2}-a x+\frac{a}{2}
$$

is always positive, the range of the real number $a$ is $\qquad$ $ـ$.
3. $(0,2]$.

Notice that, the graph of the quadratic function $y=x^{2}-a x+\frac{a}{2}$ is a parabola, with its axis of symmetry at $x=\frac{a}{2}$.
When $-10 \Leftrightarrow 0<a<2$;
When $a \geqslant 2$, $f(x)$ is a decreasing function in the interval $(-1,1)$,
The problem statement $\Leftrightarrow f(1)=1-\frac{a}{2} \geqslant 0 \Leftrightarrow a=2$;
When $a \leqslant-2$, $f(x)$ is an increasing function in the interval $(-1,1)$,

The problem statement $\Leftrightarrow f(-1)=1+\frac{3 a}{2} \geqslant 0$, which contradicts $a \leqslant-2$.
In summary, the range of values for $a$ is $(0,2]$.