SO(2) is a continuous abelian group describing proper rotations in 2d space. Because all elements commute, each conjugacy class contains only one element. Thus, there are uncountably many classes that can be parameterized by the angle $0\le\phi<2\pi$. According to the group theory, the number of classes is equal to the number of irreducible representations (irreps). Thus, there should be uncountably many irreps. However, the irreps are known. They can be parametrized by integer (or half-integer) $J$, which has the meaning of the angular momentum in quantum mechanics. There are countably many of them. What are the remaining uncountably many representations? Am I missing something in these considerations?
How many irreducible representations does SO(2) have
lie-groupsrepresentation-theory
Related Solutions
The conjugacy classes you have provided correspond to partitions of $5$ by taking cycle types. The partitions of $5$ naturally index the irreducible representations of $S_5$. The irreducible representation of $S_5$ corresponding to a partition $\lambda$ is called the Specht module $S^{\lambda}$. These representations can be described explicitly. A reference is "Young Tableaux" by Fulton.
The characters of Specht modules are reasonably computable, at least for $S_5$. There is the Jacobi-Trudi formula expressing the Schur polynomial $s_{\lambda}$ in terms of the polynomials $h_{\mu}$. Using the isomorphism between the representation ring $R=\bigoplus_{n=1}^{\infty}R_n$ ($R_n$ being the representation ring of $S_n$) and the ring $\Lambda$ of symmetric functions, you obtain a corresponding formula for the character of $S^{\lambda}$.
For your other question, a one-dimensional representation $\varphi:G\rightarrow\mathbb{C}^*$ of a finite group $G$ always factors through the abelianization $G/[G,G]$ of $G$. (Here, $[G,G]$ is the commutator subgroup of $G$.) So, you can reduce to the case of finding all one-dimensional representations $\psi:G/[G,G]\rightarrow\mathbb{C}^*$. In your case, the commutator subgroup of $S_5$ is $A_5$, so that the abelianization is isomorphic to $\mathbb{Z}_2$. The latter group has exactly two one-dimensional representations up to isomorphism.
Here is one way to get the conjugacy classes of $D_n$ and irreducible representations over $\mathbf{C}$.
Setup
First I will fix some notation. A presentation of $D_n$ is $\langle r, s\mid r^n = s^2 = 1, sr=r^{-1}s\rangle$, which means we can pin down $D_n$ as a group of rotations $\{ 1, r, r^2\ldots r^{n-1}\}$ together with a bunch of reflections $\{ s, sr, sr^2\ldots sr^{n-1}\}$. (In what follows I will frequently be sloppy and implicitly use the identifications $r^{-i} = (r^i)^{-1} = r^{n-i}$.)
Analysis of conjugacy classes of $D_n$
To figure out the conjugacy classes, we can just use brute force. Every element is $r^i$ or $sr^i$ for $0\leq i < n$, so it's not too hard just to write down every possible conjugation:
\begin{array}{rlclcl} \text{Conjugate} &r^i &\text{by}&r^j &:&(r^j) r^i(r^{-j})=r^i\\ &r^i &\text{by}&sr^j&:&(sr^j)r^i(r^{-j}s)= sr^is=r^{-i}\\ &sr^i&\text{by}&r^j &:&(r^j) sr^i(r^{-j}) = sr^{-j}r^ir^{-j} = sr^{i-2j}\\ &sr^i&\text{by}&sr^j&:&(sr^j)sr^i(r^{-j}s) = r^{i-2j}s=sr^{2j-i}\\ \end{array}
Conjugacy classes of rotations
The first two rows tell us that the set of rotations decomposes into inverse pairs $r^i$ and $(r^{i})^{-1}$, i.e. the classes $\{1\}, \{r, r^{n-1}\}, \{r^2, r^{n-2}\},\ldots$
Counting these, there are $\frac{n}{2}+1$ when $n$ is even (note that $r^{n/2}$ is its own inverse) and $\frac{n+1}{2}$ when $n$ is odd.
Conjugacy classes of reflections
Now observe from the third and fourth lines of the table that $sr$ is conjugate to $sr^3, sr^5, \ldots$ while $s$ is conjugate to $sr^2, sr^4,\ldots$ and these two sets are disjoint if $n$ is even. However, $sr$ is conjugate to $sr^{n-1}$ (via $r$) so if $n$ is odd, all the nontrivial reflections are in one conjugacy class. (You said you already knew this but I'm putting it here for completeness.)
Together, this brings us to the total number of conjugacy classes of $D_n$: \begin{array}{rl} \left(\frac{n}{2}+1\right)+2 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\ \left(\frac{n+1}{2}\right)+1 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.} \end{array}
Analysis of irreducible representations of $D_n$
One dimensional irreducibles
The commutators of $D_n$ look like $[r^i, sr^j]$ or the inverse of such, and
$$[r^i, sr^j] = r^{-i}(sr^j)r^i(sr^j) = sr^{2i+j}sr^j = (r^i)^2$$ so the commutators generate the subgroup of squares of rotations. This means $G/[G,G]$ has order 2 if $n$ is odd (since all rotations are squares) or order 4 if $n$ is even (since only half the rotations are squares). Now you can use your fact #4, which tells us that we have precisely 2 ($n$ odd) or 4 ($n$ even) irreps of dim. 1 obtained from pulling back those from $G/[G,G]$.
Other irreducibles
This is related to your item #5. We can define some 2-dim'l representations over $\mathbf{R}$, namely
\begin{array}{ccc} r&\mapsto&\pmatrix{\cos(2\pi k/n)&-\sin(2\pi k/n)\\ \sin(2\pi k/n)&\cos(2\pi k/n)} \\ s&\mapsto&\pmatrix{0&1\\1&0} \end{array} for $0\leq k \leq \lfloor \frac{n}{2}\rfloor$. We would like to know if these rep'ns are irreducible if we consider them as matrices over $\mathbf{C}$.
They are reducible if $k=0$ or $k = n/2$ (can you decompose them?). If $k$ is different from $0$ or $n/2$, a quick computation shows that the matrix for $r$ has distinct complex eigenvalues $\pm e^{2\pi ki/n}$ with corresponding eigenvectors $\pmatrix{1\\-i}$ and $\pmatrix{1\\i}$. The spans of each e-vector are the only candidates for invariant subspaces, but the matrix for $s$ interchanges the two eigenspaces, so there are no invariant subspaces and thus these repn's are irreducible.
Final count
We have 2-dim'l irreps for each integer $1\leq k < \frac{n}{2}$, specifically we have $\frac{n}{2}-1$ for $n$ even and $\frac{n-1}{2}$ for $n$ odd. If we count these with the 1-dim'l irreps, we have
\begin{array}{rl} \left(\frac{n}{2}-1\right)+4 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\ \left(\frac{n-1}{2}\right)+2 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.} \end{array}
which matches up with the number of conjugacy classes, so we must be done by your fact #1. (Also, we can verify that the sum of squares of the dimensions of the irreps is $2n$ in both cases.)
Best Answer
The statement “the number of classes is equal to the number of irreducible representations” holds for finite groups, but not in general. And $SO(2,\mathbb R)$ is infinite.