Помогите решить со 2 по 8,во 2 нужно упростить

[tex]2.\;5^{\log_2+\log_{25}64}=5^{\log_28}\cdot5^{\log_{25}64}=5^3\cdot5^{\log_{5^2}64}=5^3\cdot5^{\frac12\log_564}=\\=5^3\cdot5^{\log_5(64)^\frac12}=5^3\cdot5^{\log_58}=5^3\cdot8=125\cdot8=1000[/tex]
[tex]3.\;\cos3\alpha\cos4\alpha+\sin3\alpha\sin4\alpha-\cos\left(\alpha-\frac\pi2\right)=\\=\frac{\cos\alpha+\cos7\alpha}{2}+\frac{\cos\alpha-\cos7\alpha}2-\cos\left(\frac\pi2-\alpha\right)=\cos\alpha-\sin\alpha\\4.\;4\cdot0,5^{x+1}-0,5^x+0,5^{x-2}=20\\4\cdot0,5\cdot0,5^x-0,5^x+0,5^{-2}\cdot0,5^x=20\\2\cdot0,5^x-0,5^x+4\cdot0,5^x=20\\5\cdot0,5^x=20\\0,5^x=4\\0,5^x=0,5^{-2}\\x=-2[/tex]
[tex]5.\;\log_{\frac9{16}(2-0,1x)}\leq\frac12\\O.D.3.:\;2-0,1x>0\Rightarrow0,1x<2\Rightarrowx<20\\2\log_{\frac9{16}}(2-0,1x)\leq1\\\log_{\frac34}(2-0,1x)\leq1\\2-0,1x\geq\frac34\\0,1x\leq\frac54\\x\leq\frac{50}4\\x\leq12,5[/tex]
[tex]6.\;y=\sqrt{\left(\frac12\right)^{2x+9}-\frac18}\\\left(\frac12\right)^{2x+9}-\frac18\geq0\\\left(\frac12\right)^{2x+9}\geq\frac18\\\left(\frac12\right)^{2x+9}\geq\left(\frac12\right)^3\\2x+9\leq3\\2x\leq-6\\x\leq-3[/tex]
[tex]7.\;\sin\left(\pi+x\right)=\cos\left(-\frac\pi3\right)\\\sin(\pi+x)=\frac12\\\pi+x=\frac\pi6+2\pi n\\x=-\frac{5\pi}6+2\pi n,\;n\in\mathbb{Z}[/tex]
[tex]8.\;y=\sqrt{5x-1}\\y'=\frac5{2\sqrt{5x-1}}\\y'(1)=\frac5{2\sqrt{5-1}}=\frac54=1\frac14=1,25[/tex]