(1)
We have problems only when parallel lines are colored in the same color.
(If in at least one direction we have two colors, then make a choice of a line making the angle $\alpha$ with it...)
So we assume this "bad case" already happens.

We separate the cases $\alpha \not\in \Bbb Q\cdot \pi$ and
$\alpha\in \Bbb Q\cdot \pi$. The first case first. The system $\{\alpha,\pi\}$ is $\Bbb Q$-linearly independent, so we can choose a basis of $\Bbb R$ over $\Bbb Q$ extending it. Each $x\in \Bbb R$ can then be written uniquely as
$$
x = x_\alpha\cdot \alpha +x_\pi\cdot \pi+\dots
$$
as a finite sum with coefficients $x_\alpha,x_\pi,\dots\in\Bbb Q$. (Only finitely many coefficients are non zero, we omit the other terms from the above linear combination to have a finite sum.)

Now we color the possible rational numbers $x_\alpha$ in two colors, so that $r\in \Bbb Q$ and $(r+1)\in \Bbb Q$ always have a different color. For instance we can do the following. Fix $q\in\Bbb N_{>0}$, a possible denominator. Fix some colors for $0/q,\dots,(q-1)/q\in[0,1)$. Then take the opposite colors for the fractions with denominator $q$ in $[1,2)$, then pass again to the opposite color in the next interval $[2,3)$, and so on, do "the same" alternatively for $[-1,0)$, $[-2,-1)$, ... Or just simply use the same color for $r$ in the intervals $[n,n+1)$, $n\in \Bbb Z$ even, and the other color for $r$ in the other intervals.

We now color the line making the angle $x\in\Bbb R$ with $Ox$ by the color associated as above to the rational number $x_\alpha$. This does not depend on adding a (rational) multiple of $\pi$ to $x$, so the coloring is well defined. However, $x$ and $x+\alpha$ always have a different coloring, since the rational numbers
$$
x_\alpha\text{ and }
(x+\alpha)_\alpha=x_\alpha+1
$$
have a different coloring.

The remained case, $\alpha=\frac pq\cdot \pi$, where the fraction $p/q$ is irreducible, $p,q\in\Bbb Z$, $q>0$.

If $q$ is even, (so $p$ is odd,) then we color the sectors delimited by the regular polygon with $q$ sides alternatively, i.e. the lines making an angle in $\left[\ k\cdot\frac {2\pi}q\ ,\ (k+1)\cdot\frac {2\pi}q\ \right)$ in the "color" $k$ modulo $2$. Then any two lines forming the angle $\alpha$ have a different color.

Consider now the case $q$ is odd. Yes, it is finally possible to realize the coloring. Consider the list of angles w.r.t. $Ox$:
$$
0\cdot\frac {2\pi}q\ ,\
p\cdot\frac {2\pi}q\ ,\
2p\cdot\frac {2\pi}q\ ,\dots\ ,\
(q-1)p\cdot\frac {2\pi}q\ ,\
qp\cdot\frac {2\pi}q\ .
$$
We have an odd number of values, the first and the last one correspond to the same color, so there are two different consecutive values in the list colored in the same way.

$\square$

(2 and 3) We have $n\ge 3$. I suppose the "color of a side" is the color of the line determined by it. We show the answer is negative, the counterexample will give the same color for parallel lines.

Fix a line / a direction $d$.
Consider the perpendicular, $d'$.
Then $d'$ and $Ox$ are forming an angle $x\in [0,\pi)$.
We use the color red for $d$ if $x\in [0,2\pi/n)$, corresponding to "the first sector" (of the unit circle) else blue.

Start now with a regular polygon with $n$ sides, $(P)=A_1A_2A_3\dots A_n$. Without loss of generality we move it in a parallel manner to be centered in the origin $O$.

It is clear, that at least one side mid point $A_{j,j+1}$ (of side $A_jA_{j+1}$ of $(P)$ is in the first sector $d_0Od_1$, say $A_{12}$, then $A_{23}$ outside this first sector, so $A_1A_2$ is red and $A_2A_3$ is blue.

We have constructed a counterexample coloring. At a closer look, only for $n\ge 4$. For $n=3$ we use the same coloring strategy for the direction $d$ itself, if it is in the first sector use white, else black.

$\square$