Лишь то что отмечено, помогите, прошу

Решите задачу:

$x+2\ \textgreater \ 0, \ x+3\ \textgreater \ 0\\ x\ \textgreater \ -2, \ x\ \textgreater \ -3\\ x\ \textgreater \ -2\\ x+2 \leq x+3\\ x-x \leq 3-2\\ 0x \leq 1\\ x\in R\\ Answer: \ x\in(-2;+\infty)\\$

$\frac{lg^2(x-3)-3lg(x-3)+3}{lg(x-3)} \geq 1\\ x-3\ \textgreater \ 0, \ lg(x-3) \neq 0\\ x\ \textgreater \ 3, \ x \neq 4\\ lg(x-3)=a\\ \frac{a^2-3a+3-a}{a} \geq 0\\ \frac{a^2-4a+3}{a} \geq 0\\ \frac{(a-1)(a-3)}{a} \geq 0\\ a\in(0;1]U[3;+\infty) \\ 0\ \textless \ lg(x-3) \leq 1, \ lg(x-3) \geq 3\\ lg1\ \textless \ lg(x-3) \leq lg10, \ lg(x-3) \geq lg1000\\ 1\ \textless \ x-3 \leq 10, \ x-3 \geq 1000\\ 1+3\ \textless \ x \leq 10+3, \ x \geq 1003\\ x\in(4;13]U[1003;+\infty)$

$log^2_{0,5}x^2+log_{0,5}x-3 \geq 0\\ x\ \textgreater \ 0\\ log_{0,5}x=a\\ 4a^2+a-3 \geq 0\\ a=-1, \ a= \frac{3}{4}\\ 4(a+1)(a- \frac{3}{4}) \geq 0\\ a\in(-\infty;-1]U( \frac{3}{4} ;+\infty) \\ log_{0,5}x \leq -1, \ log_{0,5}x \geq \frac{3}{4}\\ log_{0,5}x \leq log_{0,5}2, \ log_{0,5}x \geq log_{0,5} \sqrt[4]{0,5^3} \\ x \geq 2, \ x \leq \sqrt[4]{0,5^3}\\ Answer:x\in (0; \sqrt[4]{0,5^3}]U[2;+\infty)$