Answer
$\dfrac{-(s+1)}{\sqrt 2 [(s+2)^2+1]}$
Work Step by Step
The Laplace transform of function are given as:
$F(s)=L[\dfrac{e^{-2t} \sin t}{\sqrt 2} - L [\dfrac{e^{-2t} \cos t}{\sqrt 2}]$
The first shifting Theorem can be expressed as:
$F(s)=\dfrac{1}{\sqrt 2} [\dfrac{1}{(s+2)^2+1} -\dfrac{1}{\sqrt 2} [\dfrac{s+2}{(s+2)^2+1}]\\=\dfrac{-(s+1)}{\sqrt 2 [(s+2)^2+1]}$
