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分析（1）求出解集即求出集合A．
（2）据B⊆A．分类讨论写出集合B，利用二次方程的判别式就B的各种情况求出m的范围．
解答解：（1）$\frac{3}{x}$＞1等价于0＜x＜3，x∈Z，故A={1，2}
（2）B∩A=B即B⊆A．集合A={1，2}的子集有ϕ、{1}、{2}、{1，2}．
当B=ϕ时，△=m2-8＜0，解得$-2\sqrt{2}＜m＜2\sqrt{2}$．
当B={1}或{2}时，$\left\{\begin{array}{l}△={m^2}-8=0\\1-m+2=0\end{array}\right.或\left\{\begin{array}{l}△={m^2}-8=0\\4-2m+2=0.\end{array}\right.$，则m无解．
当B={1，2}时，$\left\{\begin{array}{l}△={m^2}-8＞0\\1+2=m\\1×2=2.\end{array}\right.⇒\left\{\begin{array}{l}m＜-2\sqrt{2}或m＞2\sqrt{2}\\m=3.\end{array}\right.⇒m=3$．
综上所述，实数m的取值范围是$-2\sqrt{2}＜m＜2\sqrt{2}$或m=3．
点评本题考查分式不等式的解法；利用集合的关系求集合；利用判别式判断二次方程根的情况．
