$(x-3)^{2x^2 -7x} > 1 = (x-3)^0$
$\Leftrightarrow \begin{cases} x-3 > 0 \\ x- 3 \ne 1 \\ (x-3 -1)(2x^2 -7x ) > 0 \end{cases} \Leftrightarrow \begin{cases} x> 3 \\x \ne 4 \\ \left [ \begin{matrix} x > 4 \\ 0 < x < \dfrac{7}{2} \end{matrix} \right. \end{cases}$
$\Leftrightarrow \left [ \begin{matrix} x > 4 \\ 3 < x < \dfrac{7}{2} \end{matrix} \right.$