Verified answer

[tex]\displaystyle\bf\\1)\\\\Cos\Big(-\frac{11\pi }{6} \Big)=Cos\Big(\frac{11\pi }{6} \Big)=Cos\Big(2\pi -\frac{\pi }{6} \Big)=Cos\frac{\pi }{6} > 0\\\\\\Cos\Big(-\frac{7\pi }{6} \Big)=Cos\Big(\frac{7\pi }{6} \Big)=Cos\Big(\pi +\frac{\pi }{6} \Big)=-Cos\frac{\pi }{6} < 0\\\\\\\boxed{Cos\Big(-\frac{11\pi }{6} \Big) > Cos\Big(-\frac{7\pi }{6} \Big)}\\\\\\2)\\\\Cos\Big(-\frac{3\pi }{2} \Big)=Cos\Big(\frac{3\pi }{2} \Big)=0[/tex]

[tex]\displaystyle\bf\\Cos\Big(-\frac{9\pi }{8} \Big)=Cos\Big(\frac{9\pi }{8} \Big)=Cos\Big(\pi +\frac{\pi }{8} \Big)=-Cos\frac{\pi }{8} < 0\\\\\\\boxed{Cos\Big(-\frac{3\pi }{2} \Big) > Cos\Big(-\frac{9\pi }{8} \Big)}\\\\\\3)\\\\Cos(-\pi )=Cos\pi =-1\\\\\\Cos\Big(-\frac{9\pi }{5} \Big)=Cos\Big(\frac{9\pi }{5} \Big)=Cos\Big(2\pi -\frac{\pi }{5} \Big)=Cos\frac{\pi }{5} > 0\\\\\\\boxed{Cos(-\pi ) < Cos\Big(-\frac{9\pi }{5} \Big)}\\\\\\4)\\\\Cos6,5\approx Cos370,5^{0} \approx Cos10,5^{0} > 0[/tex]

[tex]\displaystyle\bf\\Cos7,5\approx Cos427,5^{0} \approx Cos67,5^{0} > 0\\\\10,5^{0} < 67,5^{0} \ \ \ \Rightarrow \ \ \ \boxed{Cos6,5 > Cos7,5}\\\\\\5)\\\\Cos\frac{\pi }{5} > 0\\\\\\Cos\frac{5\pi }{4} =Cos\Big(\pi +\frac{\pi }{4} \Big)=-Cos\frac{\pi }{4} < 0\\\\\\\boxed{Cos\frac{\pi }{5} > Cos\frac{5\pi }{4} }[/tex]