Первое: (4/9)^x<=sqrt(2/3)</p>
2x>=1/2
x>=1/4
Второе: замена t=log3(x), t!=0
(t-7)/(1/t-3)<=2</p>
(7t-t^2)/(3t-1)<=2</p>
(t^2-t-2)/(3t-1)>=0
(t-2)(t+1)/(3t-1)>=0
t in [-1, 0) U (0, 1/3)U[2,+infty)
x in [1/3, 1) U (1, 3^(1/3)) U [9, +infty)
Пересекаем с решением первого, в итоге имеем [1/3, 1) U (1, 3^(1/3)) U [9, +infty)