Помогите пожалуйста решить 4 и 5 задания

Question 1

Question 2

Решите задачу:

$\log_{x+1}(x+3)\ \textless \ 1, \\ \left\{\begin{array}{c}x+3\ \textgreater \ 0,\\ \left[\begin{array}{c}\left \{ {{0\ \textless \ x+1\ \textless \ 1,} \atop {x+3\ \textgreater \ x+1,}} \right.\\ \left \{ {{x+1\ \textgreater \ 1,} \atop {x+3\ \textless \ x+1,}} \right.\end{array}\right. \end{array}\right. \left\{\begin{array}{c}x\ \textgreater \ -3,\\ \left[\begin{array}{c}\left \{ {{-1\ \textless \ x\ \textless \ 0,} \atop {0\cdot x\ \textgreater \ -2,}} \right.\\ \left \{ {{x\ \textgreater \ 0,} \atop {0\cdot x\ \textless \ -2,}} \right.\end{array}\right. \end{array}\right.$
$\left\{\begin{array}{c}x\ \textgreater \ -3,\\ \left[\begin{array}{c}\left \{ {{-1\ \textless \ x\ \textless \ 0,} \atop {x\in R,}} \right.\\ \left \{ {{x\ \textgreater \ 0,} \atop {x\in\varnothing,}} \right.\end{array}\right. \end{array}\right. \left\{\begin{array}{c}x\ \textgreater \ -3,\\ \left[\begin{array}{c} -1\ \textless \ x\ \textless \ 0, \\ x\in\varnothing, \end{array}\right. \end{array}\right. \left \{ {{x\ \textgreater \ -3} \atop {-1\ \textless \ x\ \textless \ 0,}} \right. \\ -1\ \textless \ x\ \textless \ 0, \\ x\in(-1;0)$

$|2+\frac{x}{9}|^{\log_3|2+\frac{x}{9}|}=81, \\ |2+\frac{x}{9}|\ \textgreater \ 0, \ 2+\frac{x}{9}\neq0, \ x \neq -18; \\ |2+\frac{x}{9}|=a, \\ a^{\log_3a}=81, \\ \log_3a^{\log_3a}=\log_33^4, \\ \log_3a\log_3a=4, \\ \log_3^2a=4, \\ \left [ {{\log_3a=-2,} \atop {\log_3a=2;}} \right. \left [ {{a=\frac{1}{9},} \atop {a=9;}} \right. \left [ {{|2+\frac{x}{9}|=\frac{1}{9},} \atop {|2+\frac{x}{9}|=9;}} \right.$
$\left[\begin{array}{c} 2+\frac{x}{9}=-\frac{1}{9}, \\ 2+\frac{x}{9}=\frac{1}{9}, \\ 2+\frac{x}{9}=-9, \\ 2+\frac{x}{9}=9, \end{array}\right. \left[\begin{array}{c} x=-19, \\ x=-17, \\ x=99, \\ x=63. \end{array}\right.$

Question 3

В самой нижней совокупности уравнений опечатка. Корень должен быть x=-99, а не x=99.