Помогите пожалуйста

Question 1

Помогите пожалуйста

Question 2

1.
ОДЗ:
$x-8\ \textgreater \ 0\\ x\ \textgreater \ 8$
$\sqrt{x-8}=13\\x-8=169 \\x=177$
2.
$16^{x} =256\\ 2^{4x}=2^8\\ 4x=8\\x=2$
3.
$( \frac{1}{19} )^{x-7}= \frac{1}{361} \\ ( \frac{1}{19} )^{x-7}= (\frac{1}{19})^2 \\ x-7=2\\ x=8$
4.
$22^{2x+38}=1\\ 22^{2x+38}=22^0\\ 2x+38=0\\ 2x=-38\\ x=-19$
5.
ОДЗ:
$x+41\ \textgreater \ 0\\x\ \textgreater \ -41\\ log_{7}(x+41)=4\\ log_{7}(x+41)=4log_{7}7\\ log_{7}(x+41)=log_{7}7^4\\ x+41=2401\\x=2360$
6.
ОДЗ:
$3x-9\ \textgreater \ 0\\3x\ \textgreater \ 9\\x\ \textgreater \ 3\\ log_{ \frac{1}{5} }(3x-9)=-4\\ log_{ \frac{1}{5} }(3x-9)=-4log_{ \frac{1}{5} }{ \frac{1}{5} }\\ log_{ \frac{1}{5} }(3x-9)=log_{ \frac{1}{5} }(\frac{1}{5})^{-4}\\ 3x-9=625\\ 3x=634\\ x= \frac{634}{3}$
7.
$sin^2x-5sinx-6=0\\ sinx=t, t \in [-1;1]\\ t^2 -5t-6=0\\ D=25+24=49\\ t_1= \frac{5+7}{2}=6 \notin [-1;1]\\ t_2=-1\\ sinx=-1\\ x=- \frac{ \pi }{2} +2 \pi k, k \in Z$
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1.
$9^{x+3} \leq 729\\ 9^{x+3} \leq 9^3 \\ x+3 \leq 3\\ x \leq 0$
2.
$( \frac{1}{13} )^{x+7}\ \textless \ \frac{1}{169} \\ ( \frac{1}{13} )^{x+7}\ \textless \ (\frac{1}{13})^2 \\ x+7\ \textgreater \ 2\\ x\ \textgreater \ -5$
3.
ОДЗ:
$x+17\ \textgreater \ 0\\x\ \textgreater \ -17\\ log_8(x+17)\ \textless \ 3\\ log_8(x+17)\ \textless \ 3log_88\\ log_8(x+17)\ \textless \ log_88^3\\ x+17\ \textless \ 512\\ x\ \textless \ 495$
ответ:
$-17 \ \textless \ x\ \textless \ 495$
4.
ОДЗ:
$x\ \textgreater \ 0\\ log_{0.9}x \geq 2\\ log_{0.9}x \geq 2log_{0.9}0.9\\ log_{0.9}x \geq log_{0.9}0.9^2\\ x \leq 0.81$
ответ:
$0\ \textless \ x \leq 0.81$