ПОМОГИТЕ, ПОЖАЛУЙСТА, ОЧЕНЬ СРОЧНО

Решите задачу:

$====== 1 ======\\\\ y^{17}:y^{5}=y^{17-5}=y^{12}\\ x^2 \cdot x^8:x=x^{2+8-1}=x^9\\ (xyz)^3=x^3\cdot y^3\cdot z^3\\\\ ====== 2 ======\\\\ \frac{a^4\cdot(a^2)^3}{(a^5)^2} = \frac{a^4\cdot a^6}{a^{10}} = a^{4+6-10}=a^0=1\\\\ ====== 3 ======\\\\ \frac{(3^2)^3\cdot(3^3)^4}{(3\cdot3^2)^6} = \frac{3^6\cdot 3^{12}}{(3^3)^6} = \frac{3^{18}}{3^{18}} = 1\\ \frac{(4^2)^3\cdot(4^5)^2}{(4\cdot4^2)^5} = \frac{4^6\cdot4^{10}}{(4^3)^5} =\frac{4^{16}}{4^{15}} =4\\\\$

$====== 4 ======\\\\ (a^3)\cdot (...)^2=4a^7\\ (...)^2= \frac{4a^7}{a^3} \\ (...)^2= 4a^4 \\ (...)=\pm \sqrt{4a^4} \\ (...)=\pm 2a^2 \\\\ (...)^2 \cdot a^{18}=a^{24}\\ (...)^2= \frac{a^{24}}{a^{18}} \\ (...)^2= a^6\\ (...)= \pm \sqrt{a^6}\\ (...)= \pm a^3\\\\ ====== 5 ======\\\\ ((p^7)^3)^2:p^4=p^{7\cdot3\cdot2-4}=p^{38}\\ (y^3)^2 \cdot (y^2)^3=y^6 \cdot y^6 = y^{6+6}=y^{12}$