It's not clear to me exactly what "$1$-dimensional dynamical system" means, so I'll stick with continuous (but not necessarily invertible) transformations of $S^1$, and with the $\omega$-limit taken in forward time.

$2$. The map $f \colon [-1,1] \to [-1,1]$ given by $f(x):=4x^3-3x$ has a dense orbit and fixes $1$ and $-1$. Take the disjoint union of two copies of $[-1,1]$ with this map defined on each. Now identify $1$ on the first interval with $1$ on the other, and identify $-1$ with the copy of $-1$ too. This creates a map of the circle with two invariant closed intervals. The "dense orbit" in each interval has $\omega$-limit set equal to the whole interval (i.e. a closed semicircle). I think that with a little care one could come up with an analytic example.

$4$. I don't have a complete answer, but here is a quick sketch to show that the limit set cannot consist only of $k$ distinct *fixed* points $z_1,\ldots,z_k$. I will show that in this case necessarily $k=1$.
Choose $\kappa>0$ such that $d(z_i,z_j)>2\kappa$ when $i \neq j$. Now choose $\delta \in (0,\kappa)$ such that $d(f(y_1),f(y_2))<\kappa$ whenever $d(y_1,y_2)<\delta$. Let $X$ denote the set of all points in $S^1$ which are not within an open $\delta$-ball of any $z_i$. Since $X$ is compact but does not intersect the limit set of $x$, there exists $N\geq 1$ such that $f^n(x)\notin X$ when $n \geq N$. In particular, for every $n \geq N$ there exists a unique $i_n \in \{1,\ldots,k\}$ such that  $d(f^n(x),z_{i_n})<\delta$. (Uniqueness holds since otherwise $2\kappa<d(z_i,z_j)\leq d(z_i,f^n(x))+d(f^n(x),z_j)<2\delta$, a contradiction.) I claim that $(i_n)_{n=N}^\infty$ is constant, which implies that the $\omega$-limit set is actually a point. Indeed, if this is not the case then $i_{n+1} \neq i_n$ for some $n$, but then $$2\kappa<d(z_{i_n},z_{i_{n+1}}) \leq d(f(z_{i_n}),f(f^n(x)))+ d(f^{n+1}(x),z_{i_{n+1}})< \kappa + \delta$$
contradicting the definition of $\delta$. It follows that the limit set is a point as claimed.

$5$. Asaf's proposed answer to 4 can be adapted to answer 5. Let $f \colon \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$ be the doubling map $f(x)=2x$, and let $x=\sum_{n=1}^\infty 2^{-n^2}$. A point is in the $\omega$-limit set of $x$ if and only if it is either $0$ or $2^{-k}$ for some integer $k$.