For the first example. The double inequality $0\leq x+y\leq 1$ means that $ \left\{ \begin{array}{c} 0\leq x+y \\ x+y\leq 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} y\geq -x \\ y\leq 1-x \end{array} \right. $ and $0\leq x-y\leq \pi $ means that $ \left\{ \begin{array}{c} 0\leq x-y \\ x-y\leq \pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} y\leq x \\ y\geq x-\pi. \end{array} \right. $ So the conditions $0\leq x+y\leq 1$ and $0\leq x-y\leq \pi $ are equivalent to the system of four inequalities $ \left\{ \begin{array}{c} y\geq -x \\ y\leq 1-x \\ y\leq x \\ y\geq x-\pi. \end{array}\tag{1} \right. $ The region $A$ is a rectangle limited by the four lines $y=-x$, $y=1-x$, $y=x$, $y=x-\pi $ (see figure).

To evaluate
$ I:=\iint_{A}e^{x+y}\sin (x-y)\;\mathrm{d}x\mathrm{d}\tag{2}y $
we may consider the rotated system of coordinates $x',y'$ with respect to the $x,y$ system, the rotation angle being $\theta =-\pi /4$, as shown in the figure. This corresponds to the following transformation of coordinates $ \begin{eqnarray*} x^{\prime } &=&x\cos \left( -\frac{\pi }{4}\right) +y\sin \left( -\frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}x-\frac{1}{2}\sqrt{2}y \\ y^{\prime } &=&-x\sin \left( -\frac{\pi }{4}\right) +y\cos \left( -\frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}x+\frac{1}{2}\sqrt{2}y, \end{eqnarray*} $ whose inverse is $ \begin{eqnarray*} x &=&x^{\prime }\cos \left( -\frac{\pi }{4}\right) -y^{\prime }\sin \left( - \frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}% y^{\prime } \\ y &=&x^{\prime }\sin \left( -\frac{\pi }{4}\right) +y^{\prime }\cos \left( - \frac{\pi }{4}\right) =-\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}% y^{\prime }. \end{eqnarray*} $ Since $\frac{\partial (x,y)}{\partial (x^{\prime },y^{\prime })}=1$, the integral $I$ is transformed into $ \begin{eqnarray*} I &=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\left( \int_{x^{\prime }=0}^{\pi \sqrt{ 2}/2}e^{\sqrt{2}y^{\prime }}\sin (\sqrt{2}x^{\prime })\mathrm{d}x^{\prime }\right) \mathrm{d}y^{\prime } \\ &=&\int_{y^{\prime }=0}^{\sqrt{2}/2}\sqrt{2}e^{y^{\prime }\sqrt{2}}\mathrm{d} y^{\prime } \\ &=&e-1,\tag{3} \end{eqnarray*} $ because $ \begin{eqnarray*} x-y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\left( - \frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }\right) =\sqrt{2 }x^{\prime } \\ x+y &=&\frac{1}{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }-\frac{1 }{2}\sqrt{2}x^{\prime }+\frac{1}{2}\sqrt{2}y^{\prime }=\sqrt{2}y^{\prime }. \end{eqnarray*} $
Alternatively we could split $A$ into three regions, a triangle ($0\le x\le 1/2$), a quadrilateral ($1/2\le x\le π/2$) and a triangle ($\pi/2\le x\le (1+\pi)/2$), and evaluate $I$ in the original variables $x,y$: $ \begin{eqnarray*} I &=&\int_{0}^{1/2}\left( \int_{-x}^{x}e^{x+y}\sin (x-y)\mathrm{d}y\right) \mathrm{d}x \\ &&+\int_{1/2}^{\pi /2}\left( \int_{-x}^{1-x}e^{x+y}\sin (x-y)\mathrm{d} y\right) \mathrm{d}x \\ &&+\int_{\pi /2}^{(1+\pi )/2}\left( \int_{x-\pi }^{1-x}e^{x+y}\sin (x-y) \mathrm{d}y\right) \mathrm{d}x. \end{eqnarray*} $
As for the second example $R$ is the semi-annulus centered at $(0,0)$ with outer radius equal to 2, inner radius 1 and $y\ge 0$. The Jacobian of the transformation of Cartesian to polar coordinates is $\frac{\partial \left( x,y\right) }{\partial \left( r,\theta \right) }=\sqrt{x^{2}+y^{2}}=r$. Hence $ \begin{eqnarray*} \iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\left( x^{2}+y^{2}\right) ^{2}} &=&\int_{r=1}^{2}\int_{\theta =0}^{\pi }\frac{1}{r^{4}}r\;\mathrm{d}r\mathrm{ d}\theta \\ &=&\int_{0}^{\pi }\left( \int_{1}^{2}\frac{1}{r^{3}}\mathrm{d}r\right) \mathrm{d} \theta \\ &=&\int_{0}^{\pi }\frac{3}{8}\mathrm{d}\theta \\ &=&\frac{3}{8}\pi. \end{eqnarray*} $