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.)
No. A quasisimple group has a faithful irreducible representation if and only if the centre is cyclic. Since there are simple groups with non-cyclic Schur multiplier (e.g., $PSL_3(4)$, which has Schur multiplier $4\times 4\times 3$), there are no faithful irreducible representations of $4^2.PSL_3(4)$.
Edit: I should say, morally the answer is 'yes', because finite simple groups have cyclic Schur multiplier. The problem is for some small groups, for example $PSL_3(4)$, $Sz(8)$, $PSU_4(3)$, this isn't true, and also, as one comment just reminded me, the orthogonal groups $\Omega_{4n}^+(q)$ for $q$ odd and $n\geq 2$. But there are only finitely many such examples, apart from the orthogonal groups.
Edit 2: Just to clarify, the situation for quasisimple groups is that they have faithful, irreducible representations if and only if they have a cyclic centre. This is not difficult to see by inducing a faithful linear character for the centre. In general, this is not sufficient. This MathOverflow post gives information about the general case.
Best Answer
Some stuff I scrounged up. First, an easier result: an element $g$ of a finite group $G$ is real if it is conjugate to $g^{-1}$, or equivalently if the character of $g$ is always real in every representation. So the following conditions are equivalent:
Such groups are called ambivalent. This is a necessary condition for all representations being real; the symmetric groups obviously satisfy it, but I'm not sure if all finite Coxeter groups do (edit: they do, see the comments). However, it turns out to be satisfied by all Weyl groups.
The stronger condition that every representation is real is called being totally orthogonal; apparently it's known (see the link) that any such group must be generated by involutions. It doesn't seem like necessary and sufficient conditions are known in general.
Finally, the Weyl groups satisfy an even stronger condition which implies total orthogonality, that every representation is realizable over $\mathbb{Q}$: this is claimed in Humphreys in Section 8.10 and the citation is to Benard's On the Schur indices of characters of the exceptional Weyl groups. Such a group must in particular satisfy the property that every character is integer-valued and this condition is called being rational.