Using the provided order of integration, we can parameterize [tex]D[/tex] by
[tex]D = \left\{(x,y) \mid -1 \le y \le 2 \text{ and } \max(y-2,-y-2) \le x \le \min(-y+2,y+2)\right\}[/tex]
where
[tex]\max(y-2,-y-2) = \begin{cases} -y-2 & \text{when } y<0 \\ y-2 & \text{when } y\ge0\end{cases}[/tex]
and
[tex]\min(-y+2,y+2) = \begin{cases} y+2 & \text{when } y<0 \\ -y+2 & \text{when } y\ge0\end{cases}[/tex]
Then we have two iterated integrals to compute,
[tex]\displaystyle \iint_D \sin((x+y)\pi) \, dA \\\\ ~~~~= \int_{-1}^0 \int_{-y-2}^{y+2} \sin((x+y)\pi) \, dx \, dy + \int_0^2 \int_{y-2}^{-y+2} \sin((x+y)\pi) \, dx\,dy[/tex]
Compute the integrals with respect to [tex]x[/tex].
[tex]\displaystyle \int_{-y-2}^{y+2} \sin((x+y)\pi) \, dx = -\cos((x+y)\pi) \bigg|_{x=-y-2}^{x=y+2} \\\\ ~~~~~~~~ = -\cos((y+2+y)\pi) + \cos((-y-2+y)\pi) \\\\ ~~~~~~~~ = 1 - \cos((y+1)2\pi)[/tex]
[tex]\displaystyle \int_{y-2}^{-y+2} \sin((x+y)\pi) \, dx = -1 + \cos((y-1)2\pi)[/tex]
Compute the remaining integrals.
[tex]\displaystyle \int_{-1}^0 \left(1 - \cos((y+1)2\pi)\right) \, dy = \left(y - \frac1{2\pi} \sin((y+1)2\pi)\right)\bigg|_{y=-1}^{y=0} \\\\ ~~~~~~~~ = \left(0 - \frac1{2\pi} \sin(2\pi)\right) - \left(-1 - \frac1{2\pi} \sin(0)\right) = 1[/tex]
[tex]\displaystyle \int_0^2 \left(-1 + \cos((y-1)2\pi)\right) \, dy = \left(-y + \frac1{2\pi} \sin((y-1)2\pi)\right)\bigg|_{y=0}^{y=2} \\\\ ~~~~~~~~ = \left(-2 + \frac1{2\pi} \sin(2\pi)\right) - \left(0 + \frac1{2\pi} \sin(-2\pi)\right) = -2[/tex]
Then the overall integral has a value of
[tex]\displaystyle \iint_D \sin((x+y)\pi) \, dA = 1 + (-2) = \boxed{-1}[/tex]
