Решим неравенства методом интервалов :

[tex]\displaystyle\bf\\1)\\\\x^{2} +6x-7 > 0\\\\(x-1)\cdot(x+7) > 0\\\\\\+ + + + + \Big(-7\Big)- - - - - \Big(1\Big)+ + + + + \\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ -7\Big)\cup\Big(1 \ ; \ +\infty\Big)\\\\2)\\\\x^{2} -7x-8\leq 0\\\\(x-8)\cdot(x+1)\leq 0\\\\\\+ + + + + \Big[-1\Big]- - - - - \Big[8\Big]+ + + + + \\\\\\Otvet \ : \ x\in\Big[-1 \ ; \ 8\Big][/tex]