$0<x<\dfrac{π}{2}\Rightarrow sinx>0\\ \Rightarrow sinx=\sqrt{1-cos^2x}=\dfrac{\sqrt{15}}{4}$
$\Rightarrow cos2x=2cos^2x-1=\dfrac{-7}{8}\\ sin2x=2sinxcosx=\dfrac{\sqrt{15}}{8}\\ \Rightarrow tan2x=\dfrac{sin2x}{cos2x}=\dfrac{-\sqrt{15}}{7}\\ cot2x=\dfrac{1}{tan2x}=\dfrac{-7}{\sqrt{15}}$
$sin(x-\dfrac{π}{4})=sinx.cos\dfrac{π}{4}-sin\dfrac{π}{4}.cosx\\ =\dfrac{\sqrt{15}}{4}.\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}.\dfrac{1}{4}=\dfrac{\sqrt{15}-1}{4\sqrt{2}}$
$cos(\dfrac{π}{3}+x)=cos\dfrac{π}{3}.cosx-sin\dfrac{π}{3}.sinx\\ =\dfrac{1}{2}.\dfrac{1}{4}-\dfrac{\sqrt{3}}{2}.\dfrac{\sqrt{15}}{4}=\dfrac{1-3\sqrt{5}}{8}$
