3. Given $f(x)=2 m x^{2}-2 m x-8 x+9, g(x)=m x-m$, for any $x \in \mathbf{R}, f(x)$ and $g(x)$ at least one is positive, then the range of real number $m$ is $\qquad$
Obviously, $m \leqslant 0$ does not meet the requirements. When $m>0$, since $x>1$ implies $g(x)>0$, we only need $f(x)>0$ for $x \leqslant 1$. And $f(1)=1>0$, for $x \leqslant 0$, $2 m x^{2}-2 m x-8 x+9>0$ is obviously true; for $0 < x < 1$, $2 m x^{2}-2 m x-8 x+9>0 \Rightarrow 2 m<\frac{9-8 x}{x-x^{2}}$ $\xlongequal{t=9-8 x \in(1,9)} \frac{64 t}{8(9-t)-(9-t)^{2}}=\frac{64}{10-\left(t+\frac{9}{t}\right)} \Rightarrow 2 m<16 \Rightarrow m<8$, so the range of real number $m$ is $(0,8)$.