The greatest value of the function  $f\left(x\right)=\frac{\sin 2x}{\sin (x+\frac{\pi }{4})}$  on the interval   $[0,\frac{\pi }{2}]$  is

$f\left(x\right)=\frac{-\cos 2(x+\frac{\pi }{4})}{\sin (x+\frac{\pi }{4})}=\frac{2{\sin }^2(x+\frac{\pi }{4})-1}{\sin (x+\frac{\pi }{4})}$

Let $\sin (x+\frac{\pi }{4})=t,t\in [\frac{1}{\sqrt{2}},1]$

$\Rightarrow f\left(t\right)=\frac{2t^2-1}{t}\Rightarrow f^{\mathrm{\prime }}\left(t\right)=2+\frac{1}{t^2}>0$

$\therefore$  $\text{maximum of }f\left(t\right) \text{occurs at }t=1$  and its value is 1.