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
$$
\begin{eqnarray*}
\iint_{R}\frac{\mathrm{d}x\mathrm{d}y}{\left( x^{2}+y^{2}\right) ^{2}}
&=&\int_{r=1}^{2}\int_{\theta =0}^{\pi }\frac{1}{r^{4}}r\;\mathrm{d}r\mathrm{
d}\theta \\
&=&\int_{0}^{\pi }\left( \int_{1}^{2}\frac{1}{r^{3}}\mathrm{d}r\right) \mathrm{d}
\theta \\
&=&\int_{0}^{\pi }\frac{3}{8}\mathrm{d}\theta \\
&=&\frac{3}{8}\pi.
\end{eqnarray*}
$$