Answer
$$e^{-t+\frac{\pi}{4}} [-\cos 2t-\sin 2t] $$
Work Step by Step
We are given that $f(t)=e^{-t} (\sin (2t)+\cos 2t) $ and $a=\dfrac{\pi}{4}$
When $a$ has a positive value then , shift the function to the right for $a$ units and when $a$ has a negative value then , shift the function to the left for $a$ units
Now, $f(t-a) =f(t-\dfrac{\pi}{4}) \\=e^{-t+\frac{\pi}{4}} [\sin 2(t-\dfrac{\pi}{4}) +\cos 2(t-\dfrac{\pi}{4}) ]\\=e^{-t+\frac{\pi}{4}} [-\cos 2t-\sin 2t] $
