$0<a<\dfrac{π}{2}\Rightarrow cosa>0$
Ta có $cos^2a+sin^2a=1$
$\Rightarrow cosa=\sqrt{1-sin^2a}=sqrt{1-\dfrac{1}{10}}=\dfrac{3}{\sqrt{10}}$
$sin2a=2sinacosa=2.\dfrac{1}{\sqrt{10}}.\dfrac{3}{\sqrt{10}}=\dfrac{3}{5}$
$tana=sina:cosa=\dfrac{1}{\sqrt{10}}:\dfrac{3}{\sqrt{10}}=\dfrac{1}{3}$
$cos(a-\dfrac{π}{6})=cosa.cos\dfrac{π}{6}+sina.sin\dfrac{π}{6}\\=\dfrac{3}{\sqrt{10}}.\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{10}}.\dfrac{1}{2}=\dfrac{3\sqrt{3}+1}{2\sqrt{10}}$
$tan(a-\dfrac{π}{4})=\dfrac{tana-tan\dfrac{π}{4}}{1+tana.tan\dfrac{π}{4}}=\dfrac{\dfrac{1}{3}-1}{1+\dfrac{1}{3}.1}=\dfrac{-1}{2}$