Помогите решить задания, заранее благодарю!

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$\frac{2^{-2}}{(2^2)^{-2}} = \frac{2^{-2}}{(2)^{-4}} =2^{-2-(-4)}=2^{-2+4}=2^2=4$

Ответ: 4)

================= 2 =================

$25x^2-81=0\\ 25x^2=81\\ x^2= \frac{81}{25}\\ x_{1,2}=\pm \sqrt{\frac{81}{25}} \\ x_{1,2}=\pm \frac{9}{5} \\ x_1\cdot x_2= \frac{9}{5} \cdot (-\frac{9}{5}) =-\frac{81}{25}$

Ответ: 2)

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$\sqrt{x-2}=8-x\\\\ \left \{ {{x-2 \geq 0} \atop {8-x \geq 0}} \right. \\\\ \left \{ {{x \geq 2} \atop {-x \geq -8}} \right. \\\\ \left \{ {{x \geq 2} \atop {x \leq 8}} \right. \\\\ x\in [2;8]\\\\ ( \sqrt{x-2})^2=(8-x)^2\\ x-2=64-16x+x^2\\ 64-16x+x^2-x+2=0\\ x^2-17x+66=0\\ D=289-264=25\\ x_1= \frac{17+5}{2} = \frac{22}{2}=11 \notin [2;8]\\ x_2= \frac{17-5}{2} = \frac{12}{2}=6\\$

Ответ 3)

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$cos30cos15-sin30sin15=cos(30+15)=cos45= \frac{ \sqrt{2} }{2}$

Ответ: 4)