Помогите!!!!!Пожалуйста!!!!Очень срочно!!!!2и 3

Решите задачу:

$1)\; \; \left \{ {{log_{1/3}(3-2x)\ \textgreater \ -1} \atop {0,3^{7+4x}\ \textgreater \ 0,027}} \right. \; \; \left \{ {{3-2x\ \textless \ ( \frac{1}{3} )^{-1}\; ,\; 3-2x\ \textgreater \ 0} \atop {0,3^{7+4x}\ \textgreater \ 0,3^3}} \right. \; \; \left \{ {{3-2x\ \textless \ 3\; ,\; x\ \textless \ \frac{3}{2}} \atop {7+4x\ \textless \ 3}} \right. \\\\ \left \{ {{0\ \textless \ x\ \textless \ \frac{3}{2}} \atop {x\ \textless \ -1}} \right. \; \; \Rightarrow \; \; \; x\in \varnothing \\\\2)\; \; f(x)= \frac{5-2x^6}{1-x^3} \\\\f'(x)= \frac{-12x^5(1-x^3)-(5-2x^6)(-3x^2)}{(1-x^3)^2} = \frac{-12x^5+12x^8+15x^2-6x^8}{(1-x^3)^2} =$

$=\frac{6x^8-12x^5+15x^2}{(1-x^3)^2}=\frac{3x^2(2x^6-4x^3+5)}{(1-x^3)^2}$

$3)\; \; f(x)= (\frac{1}{5}x-9)^ {15}+ \sqrt{4x+5} \\\\f'(x)=15\cdot ( \frac{1}{5}x-9)^4\cdot \frac{1}{5} +\frac{1}{2\sqrt{4x+5}}\cdot 4=3\cdot (\frac{x}{5}-9)^4+\frac{2}{\sqrt{4x+5}}$