Answer
$$y =-x+2$$
Work Step by Step
Given $$y=(x+2)e^{-x},\ \ \ (0,2)$$
Since
$$y'= -(x+2)e^{-x}+e^{-x}=-( x+1)e^{-x}$$
Then $m=y'\bigg|_{(0,2)}=-1$, hence the tangent line given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-2}{x-0}&=-1\\
y&=-x+2
\end{align*}