Well, both the $\Omega$s are okay!
About how to depict the first one, I thought it like this:
Thinking $z=x+iy$, then $|Im(z)|<\frac{\pi}{2}$ is saying $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Now, $$f(z)=e^{(iz)}=e^{i(x+iy)}=e^{ix-y}=e^{-y}e^{ix}=e^{-y}\left(Cos(x)+iSin(x)\right)$$ And if you think about the last two expressions, they are just a circle $(e^{ix})$ with radius $e^{-y}$. Of course, $y$ is not a constant, and therefore it is a changing radius, but it does have boundaries; since:
$$-\frac{\pi}{2}<y<\frac{\pi}{2}$$ $$\frac{\pi}{2}>-y>-\frac{\pi}{2}$$ $$0.2079\approx e^{-\frac{\pi}{2}}<e^{-y}<e^{\frac{\pi}{2}}\approx 4.81$$ So in the end you will have an area trapped between a circle of radius $\approx0.2079$ and a bigger one of radius $\approx4.81$
Like this:
As for the second one, I would parametrize $U(1,2)$ as $\left(x(t), y(t)\right)=(r(Cos(t)+1),rSin(t))$ where $0\le r<2$ , $0\le t \le2\pi$ and replace that into $f(z)$ for the $x$ and $y$s. First separate $f(z)$ into its complex and imaginary parts: $$f(z)=\frac{2z-1}{z+3}\frac{z+3}{z+3}=\frac{2x^2-2y^2-6x+3}{x^2-y^2-9}-6yi$$
That will give you a parametrization for the $(u,v)$ plane. I tried simplifying it but didn't succeed. An easy way is plotting the case where $r=2$ and testing where one point from the $U(1,2)$ goes, and then you can see which area corresponds to the domain $\Omega$ you are taking.
I really hope this helps!
EDIT:
This is the parametrized plot for $r=2$
I just realized this second function is a Möbius transformation of the circle. That is another, simpler way of seeing how it is transformed.