For $\Omega_S$, it might be easiest to perform by inspection. For a vector pointing to a point on the circle, draw the normal vector on the circle. You should be able to see that the vectors have the same length and direction, i.e., $n(x) = x$.
For $\Omega_C$, the boundary of the triangle is made up of three lines: $x_2=x_1$, $x_2=-x_1$, and $x_2=1$. We can easily calculate the normal vector by thinking about the perpendicular lines for these. For instance, $x_2=x_1$ has a slope of $1$ so the perpendicular line will have a slope of $-1$ (for $y=mx$ the perpendicular line will have a slope of $\frac {-1}m$). We wish to pick a vector parallel to $x_2 = (-1)\cdot x_1$ and in the correct direction so that if points outward from the boundary, e.g. $(1,-1)$. To make it the unit normal, we need the vector to have length one. Normalizing $(1,-1)$ gives $\frac 1 {\sqrt 2} (1,-1)$. You can do the same thing for $x_2 =-x_1$, and the normal vector for $x_2 = 1$ is easy to find.
For $\Omega_p$, you can notice that the boundary lies on the parabola $x_2 = x_1^2$. Thus $r(x_1) = (x_1, x_1^2)\in\mathbb R^2$ parameterizes the boundary for $x_1 \in(-1,1)$. The taking the derivative of $r$ give the tangent vector. The normalized tangent vector at $(x_1, x_2)$ is $$T(x_1) =\frac{r'(x_1)}{\|r'(x_1)\|} = \frac {(1, 2x_1)} {\sqrt{1+4x_1^2}} = \left(\frac 1 {\sqrt{1+4x_1^2}},
\frac {2x_1}{\sqrt{1+4x_1^2}}\right).$$ Now for any vector in $(a,b)\in\mathbb R^2$, we have that $(b,-a)$ is perpendicular (this is most easily seen by taking the dot product $(a,b)\cdot(b,-a)=ab - ab =0$). So $$\left(\frac {2x_1}{\sqrt{1+4x_1^2}}, \frac {-1} {\sqrt{1+4x_1^2}}\right)$$ is perpendicular to the tangent vector $T(x_1)$ at $(x_1,x_2)$ so it must be parallel to the normal vector at $(x_1, x_2)$. We can see that the above vector has length one and is pointing out from the region, so it must be the unit normal vector.
