[tex]\displaystyle\bf\\1)\\\\(x-3)(2x+3) > -9\\\\2x^{2} +3x-6x-9+9 > 0\\\\2x^{2} -3x > 0 \ |:2\\\\x^{2} -1,5x > 0\\\\x(x-1,5) > 0\\\\+ + + + + (0)- - - - - (1,5)+ + + + + \\\\Otvet: \ x\in(-\infty \ ; \ 0) \ \cup \ (1,5 \ ; \ +\infty)[/tex]

[tex]\displaystyle\bf\\2)\\\\\frac{-8}{x^{2} +2x-15} \leq 0\\\\\\8\cdot(x+5)(x-3) > 0\\\\(x+5)(x-3) > 0\\\\+ + + + + (-5)- - - - - (3)+ + + + + \\\\Otvet: \ x\in(-\infty \ ; \ -5) \cup \ (3 \ ; \ +\infty)[/tex]

[tex]\displaystyle\bf\\\left \{ {{8x+2y=-6} \atop {-y-x^{2} =6}} \right. \\\\\\\left \{ {{4x+y=-3} \atop {y=-x^{2}-6 }} \right. \\\\\\\left \{ {{4x-x^{2} -6=-3} \atop {y=-x^{2} -6}} \right. \\\\\\\left \{ {{x^{2} -4x+3=0} \atop {y=-x^{2} -6}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1}=1 \\x_{2} =3\end{array}\right} \atop {y=-x^{2} -6}} \right.[/tex]

[tex]\displaystyle\bf\\\left[\begin{array}{ccc}\left \{ {{x_{1} =1} \atop {y_{1} =-1-6}} \right. \\\left \{ {{x_{2} =3} \atop {y_{2} =-3^{2} -6}} \right. \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =1} \atop {y_{1} =-7}} \right. \\\left \{ {{x_{2} =3} \atop {y_{2} =-15}} \right. \end{array}\right\\\\\\Otvet: \ (1 \ ; \ -7) \ \ , \ \ (3 \ ; -15)[/tex]