P命题可化为-2<=1-(x-1)/3<=2,化简后可得-2<=x<=10,所以非P为x>10或x<-2.
Q命题：先十字交叉因式分解,把原不等式化为[x-(1-m)][x-(1+m)]<=0,因为m>0,所以1-m<1+m,所以1-m<=x<=1+m.非Q为x>1+m或x<1-m.
非P是非q的必要不充分条件,所以1-m<=-2或1+m>=10,所以m>=3