Ответ:

[tex]\displaystyle (-1;+\infty)[/tex]

Объяснение:

[tex]\displaystyle(tg\frac{\pi}{3})^{x-1} > 9^{-0,5}\\\\(\sqrt {3})^{x-1} > (3^2)^{-0,5}\\\\ (3^{\frac{1}{2}})^{x-1} > 3^{-1}\\\\3^{\frac{1}{2}(x-1)} > 3^{-1}\\\\\frac{1}{2}(x-1) > -1\ \ \ |\cdot 2\\\\x-1 > -2\\\\x > -2+1\\\\x > -1\\\\x\in(-1;+\infty)[/tex]