Ответ:

[tex](-\infty;-\frac{1}{3})\cup(\frac{1}{2};+\infty)[/tex]

Пошаговое объяснение:

[tex]a\neq \frac{1}{2}\\\\(2a-1)x+5=0\\\\(2a-1)x=-5\ \ \ |:(2a-1)\\\\x=-\frac{5}{2a-1}\\\\-\frac{5}{2a-1} < 3\\\\-\frac{5}{2a-1}-3 < 0\\\\-\frac{5}{2a-1}-\frac{3(2a-1)}{2a-1} < 0\\\\\frac{-5-3(2a-1)}{2a-1} < 0\\\\\frac{-5-6a+3}{2a-1} < 0\\\\\frac{-6a-2}{2a-1} < 0\\\\(-6a-2)(2a-1) < 0\\\\-6(a+\frac{1}{3})\cdot 2(a-\frac{1}{2}) < 0\\\\-12(a+\frac{1}{3})(a-\frac{1}{2}) < 0\ \ \ |:(-12)\\\\(a+\frac{1}{3})(a-\frac{1}{2}) > 0\\\\a\in(-\infty;-\frac{1}{3})\cup(\frac{1}{2};+\infty)[/tex]