Помогите решить неравенство

Question 1

Question 2

Решите задачу:

$1)\quad 25\cdot 0,04^{2x}\ \textgreater \ 0,2^{x(3-x)}\\\\25\cdot ( \frac{1}{25} )^{2x}\ \textgreater \ (\frac{1}{5})^{3x-x^2}\\\\5^{2-4x}\ \textgreater \ 5^{x^2-3x}\\\\2-4x\ \textgreater \ x^2-3x\\\\x^2+x-2\ \textless \ 0\\\\(x-1)(x+2)\ \textless \ 0\qquad +++(-2)---(1)+++\\\\x\in (-2,1)$

$2)\quad 4^{x}-2^{x}\ \textless \ 12\\\\2^{2x}-2^{x}-12\ \textless \ 0\\\\t=2^{x}\ \textgreater \ 0\; ,\; t^2-t-12\ \textless \ 0\; ,\; \; t_1=-3,\; t_2=4\\\\(t+3)(t-4)\ \textless \ 0\qquad +++(-3)---(4)+++\\\\-3\ \textless \ t\ \textless \ 4\; \; \; \to \quad 0\ \textless \ 2^{x}\ \textless \ 4\\\\2^{x}\ \textless \ 2^2\\\\x\ \textless \ 2\\\\x\in (-\infty ,2)$

$3)\quad ( \frac{1}{4})^{x}-32\cdot (\frac{1}{8})^{x^2-1}\ \textless \ 0\\\\2^{-2x}-2^5\cdot 2^{-3(x^2-1)}\ \textless \ 0\\\\2^{-2x}-2^{-3x^2+8}\ \textless \ 0\\\\2^{-2x}\ \textless \ 2^{-3x^2+8}\\\\-2x\ \textless \ -3x^2+8\\\\3x^2-2x-8\ \textless \ 0\\\\D/4=25,\; \; x_1=-\frac{4}{3}\; ,\; \; x_2=2\\\\3(x+\frac{4}{3})(x-2)\ \textless \ 0\qquad +++(- \frac{4}{3} )---(2)+++\\\\x\in (- \frac{4}{3} ,2)$

$4)\quad 3\cdot 9^{x}+11\cdot 3^{x}\ \textless \ 4\\\\3\cdot 3^{2x}+11\cdot 3^{x}-4\ \textless \ 0\\\\t=3^{x}\ \textgreater \ 0\; ,\; \; 3t^2+11t-4\ \textless \ 0\\\\D=169\; ,\; \; t_1=-4,\; \; t_2=\frac{1}{3}\\\\3(t+4)(t-\frac{1}{3})\ \textless \ 0\qquad +++(-4)---(\frac{1}{3})+++\\\\-4\ \textless \ t\ \textless \ \frac{1}{3}\quad \to \quad 0\ \textless \ 3^{x}\ \textless \ \frac{1}{3}\\\\3^{x}\ \textless \ 3^{-1}\\\\x\ \textless \ -1\\\\x\in (-\infty ,-1)$