Ответ:
0\end{array}\right\ \ \left\{\begin{array}{ccc}x>-2,5\\x\in (-4;4)\end{array}\right\ \ \to \ \ x\in (-2,5\, ;\, 4\, )\\\\\\log_{0,2}\dfrac{2x+5}{16-x^2}0,2\ \ ,\ \ \ \dfrac{2x+5-3,2+0,2x^2}{(4-x)(4+x)}>0\ ,\\\\\\\dfrac{0,2x^2+2x+1,8}{(4-x)(4+x)}>0\ \ ,\ \ \dfrac{(x+9)(x+1)}{(x-4)(x+4)}" alt="5)\ \ log_{0,2}(2x+5)-log_{0,2}(16-x^2)0\\16-x^2>0\end{array}\right\ \ \left\{\begin{array}{ccc}x>-2,5\\x\in (-4;4)\end{array}\right\ \ \to \ \ x\in (-2,5\, ;\, 4\, )\\\\\\log_{0,2}\dfrac{2x+5}{16-x^2}0,2\ \ ,\ \ \ \dfrac{2x+5-3,2+0,2x^2}{(4-x)(4+x)}>0\ ,\\\\\\\dfrac{0,2x^2+2x+1,8}{(4-x)(4+x)}>0\ \ ,\ \ \dfrac{(x+9)(x+1)}{(x-4)(x+4)}" align="absmiddle" class="latex-formula">
1\ \ ,\ \ ODZ:\ \ x^2-4x+13>0\ \to \ x\in R\\\\x^2-4x+13>10\ \ ,\ \ x^2-4x+3>0\ \ ,\ \ x_1=1\ ,\ x_2=3\\\\(x-1)(x-3)>0\ \ \ \ \ +++(1)---(3)+++\\\\x\in (-\infty ;1)\cup (3;+\infty )" alt="6)\ \ lg(x^2-4x+13)>1\ \ ,\ \ ODZ:\ \ x^2-4x+13>0\ \to \ x\in R\\\\x^2-4x+13>10\ \ ,\ \ x^2-4x+3>0\ \ ,\ \ x_1=1\ ,\ x_2=3\\\\(x-1)(x-3)>0\ \ \ \ \ +++(1)---(3)+++\\\\x\in (-\infty ;1)\cup (3;+\infty )" align="absmiddle" class="latex-formula">
-1\\x\in (-\infty ;-2\, ]\cup [\, 2\, ;+\infty )\end{array}\right\\\\\\x\in [\, 2\, ;+\infty )" alt="7)\ \ log_{0,5}(1+x-\sqrt{x^2-4})0\\x^2-4\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x^2-4}-2,5\\x>-1\\x\in (-\infty ;-2\, ]\cup [\, 2\, ;+\infty )\end{array}\right\\\\\\x\in [\, 2\, ;+\infty )" align="absmiddle" class="latex-formula">
1\ \ \to \ \ \ \sqrt{x^2-4}0\\x^2-40\\-4" alt="1+x-\sqrt{x^2-4}>1\ \ \to \ \ \ \sqrt{x^2-4}0\\x^2-40\\-4" align="absmiddle" class="latex-formula">