Let $S^2:=\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2+z^2=1\}$ be the unit sphere, $\textbf{n}:=(0,0,1)$ the northpole of $S^2$ and $\textbf{s}:=(0,0,-1)$ the southpole of $S^2$. Let $F:\mathbb{R}^2\rightarrow S^2$ ($G:\mathbb{R}^2\rightarrow S^2$ respectively) be the function that maps a point $(x,y)\in \mathbb{R}^2$ to the unique intersection point between the sphere $S^2$ and the line that passes through the point $(x,y,0)\in \mathbb{R}^3$ and the northpole $\textbf{n}\in S^2$ (the south pole $\textbf{s}\in S^2$ respectively).
a) Show that $F(\mathbb{R}^2)=S^2\setminus \{\textbf{n}\}$ and $G(\mathbb{R}^2)=S^2\setminus \{\textbf{s}\}$. Give the formulas for $F:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{n}\}$ and $G:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{s}\}$.
b) Show that the functions $F:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{n}\}$ and $G:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{s}\}$ are invertible. Give the formulas for $F^{-1}: S^2\setminus \{\textbf{n}\} \rightarrow \mathbb{R}^2$ and $G^{-1}: S^2\setminus \{\textbf{s}\} \rightarrow \mathbb{R}^2$.
c) Show that $F:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{n}\}$ and $G:\mathbb{R}^2\rightarrow S^2\setminus \{\textbf{s}\}$ define parametrizations of $S^2$, i.e. these functions are homomorphisms and embeedings.
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Do we get that $F(\mathbb{R}^2)=S^2\setminus \{\textbf{n}\}$ because $F$ is the intersection points between the sphere and the line through the north pole and so it cannot be mapped to the north pole? Or how do we get the range? Or do we have to find first the formula of $F$ ?
The function $F$ is the inverse of the stereographic projection from $S^2\setminus \{\textbf{n}\}$ to $\mathbb R^2$. There are many questions concerning stereographic projection in this forum, just make a search.
The intuition is that each $\xi = (x,y,z) \in S^2\setminus \{\textbf{n}\}$ determines a unique line $L(\xi)$ containing $\mathbf n$ and $\xi$; this line is not parallel to the $x$-$y$-plane $P$ and thus intersects $P$ is a unique point $\pi(\xi)$.
$L(\xi)$ is a parameterized by $\textbf{n} + t(\xi -\textbf{n})$, $t \in \mathbb R$. The point $\pi(\xi)$ has $z$-coordinate $0$, thus we have to solve $$1 + t(z - 1) = 0$$ which gives $t = \frac{1}{1-z}$. Note that $z < 1$ since $\xi \ne \textbf{n}$. Thus $$\pi(\xi) = \left(\frac{x}{1-z},\frac{y}{1-z},0\right) .$$ This gives the (clearly continuous) function $$\phi : S^2\setminus \{\textbf{n}\} \to \mathbb R^2, \phi(x,y,z) = \left(\frac{x}{1-z},\frac{y}{1-z}\right) .$$ You can do it also the other way as sketched in your question to get $$F : \mathbb R^2 \to S^2\setminus \{\textbf{n}\}, F(x,y) = \left(\frac{2x}{x^2+y^2+1}, \frac{2y}{x^2+y^2+1}, \frac{x^2+y^2-1}{x^2+y^2+1}\right) .$$