For Q2 $n=4$ is not possible.
Let $c(\mathbf{x}) \in \{A,B,C,D\}$ denote the color of a point $\mathbf{x}$.
As you noted in the comments, a 4-coloring with every unit square having four different-colored vertices would require
C1: No pair of points separated by a distance of $1$ or $\sqrt 2$ can have the same color.
From C1 we can show:
C2: For any point $\mathbf{o}$ and any orthonormal vectors $\mathbf{u},\mathbf{v}$ either
$$
c(\mathbf{x}+2\mathbf{u}) = c(\mathbf{x}) \text{ for every } \mathbf{x}=\mathbf{o}+m\mathbf{u}+n\mathbf{v}
$$
or
$$
c(\mathbf{x}+2\mathbf{v}) = c(\mathbf{x}) \text{ for every } \mathbf{x}=\mathbf{o}+m\mathbf{u}+n\mathbf{v}
$$
for $m,n\in\mathbb{Z}$. That is, $c$ is periodic with period $2\mathbf{u}$ or $2\mathbf{v}$ on the lattice generated by $\mathbf{u},\mathbf{v}$.
From C1 and C2 we get:
C3: For $\mathbf{o},\mathbf{u},\mathbf{v}$ as above there is some $\mathbf{x}=\mathbf{o}+m\mathbf{u}+n\mathbf{v}$ with
$$
c(\mathbf{x}) = c(\mathbf{x}+2\mathbf{u}) = c(\mathbf{x}+2\mathbf{v})
$$
or
$$
c(\mathbf{x}) = c(\mathbf{x}+4\mathbf{u}) = c(\mathbf{x}+4\mathbf{v})
$$
And C2 and C3 contradict C1, hence no such coloring is possible.
Proof: C1$\implies$C2:
Either $c(\mathbf{o}+2m\mathbf{u})=c(\mathbf{o}),c(\mathbf{o}+(2m+1)\mathbf{u})=c(\mathbf{o}+\mathbf{u}) $ for all $m$, or else there is some $\mathbf{x} = \mathbf{o}+m_0\mathbf{u}$ with $c(\mathbf{x}),c(\mathbf{x}+\mathbf{u}),c(\mathbf{x}+2\mathbf{u})$ three different colors, wlog $A,B,C$. In that case the constraints of C1 uniquely determine $c(\mathbf{x}+n\mathbf{v})$. Specifically $c(\mathbf{x}+2n\mathbf{v})=A$ and $c(\mathbf{x}+(2n+1)\mathbf{v})=C$. In a picture, starting with $\mathbf{x}$ in the top left with $\mathbf{u}$ going right and $\mathbf{v}$ going down:
$$
\begin{array}{cccc}
\cdots & A & B & C & \cdots \\
\cdots & C & D & A & \cdots \\
\cdots & A & B & C & \cdots \\
\cdots & C & D & A & \cdots \\
\cdots & A & B & C & \cdots \\
& \vdots & \vdots & \vdots
\end{array}
$$
Hence either the row containing $\mathbf{o}$ or the column containing $\mathbf{x}$ is periodic with period $2\mathbf{u}$ or $2\mathbf{v}$ respectively, and it's easy to see that either implies that the rest of the lattice is periodic with the same period.
C1 and C2 $\implies$ C3:
From C2 wlog assume coloring on the $\mathbf{o},\mathbf{u},\mathbf{v}$ lattice has period $2\mathbf{u}$. Either there is some $\mathbf{x}$ with
$c(\mathbf{x}+2\mathbf{v})=c(\mathbf{x})=c(\mathbf{x}+2\mathbf{u})$, satisfying the first condition of C3, or else $c(\mathbf{x}+2\mathbf{v})\ne c(\mathbf{x})$ for every $\mathbf{x}$ in the lattice. In that case wlog let the colors of
$\mathbf{o},\mathbf{o}+\mathbf{u},\mathbf{o}+\mathbf{v},\mathbf{o}+\mathbf{u}+\mathbf{v}$ be $A,B,C,D$ respectively, and this condition along with the constraints of C1 can fill out the coloring on the lattice. In a picture, we must continue:
$$
\begin{array}{cccc}
\cdots & A & B & A & \cdots \\
\cdots & C & D & C & \cdots \\
\cdots & B & A & B & \cdots \\
\cdots & D & C & D & \cdots \\
\cdots & A & B & A & \cdots \\
& \vdots & \vdots & \vdots
\end{array}
$$
Hence $c(\mathbf{o}+4\mathbf{v}) = c(\mathbf{o}) = c(\mathbf{o}+4\mathbf{u})$.
C2 and C3 $\implies$ not C1:
On a lattice with $\mathbf{o},\mathbf{u},\mathbf{v}$ as above, from C3 there is an $\mathbf{X}$ with $\mathbf{Y}=\mathbf{X}+t\mathbf{u}, \mathbf{Z}=\mathbf{X}+t\mathbf{v}$ and
$$ c(\mathbf{X}) = c(\mathbf{Y}) = c(\mathbf{Z}) $$
for either $t=2$ or $t=4$. There are orthonormal vectors $\mathbf{u}',\mathbf{v}'$ such that $\mathbf{Y}' = \mathbf{X}+t\mathbf{u}', \mathbf{Z}' = \mathbf{X}+t\mathbf{v}'$ and $\left\vert \mathbf{YY}'\right\vert = \left\vert \mathbf{ZZ}'\right\vert = 1$.

Then by C2 either $c(\mathbf{Y}')=c(\mathbf{X})=c(\mathbf{Y})$ or $c(\mathbf{Z}')=c(\mathbf{X})=c(\mathbf{Z})$, contradicting C1.