= Solution
For $a<b$, give the pair $\{a,b\}$ the color determined by $v_2(b-a)\bmod2$, where $v_2$ is the <2-adic valuation>: use red for even valuation and blue for odd valuation. This <finite coloring> works because the pairs $\{a,a+d\}$ and $\{a,a+2d\}$ have valuations $v_2(d)$ and $v_2(d)+1$. They therefore have opposite colors. In particular, \b[no three-term <arithmetic progression> has all its pairs <monochromatic>].
For the second assertion, $m=1$ is immediate: every singleton is a blue complete <arithmetic progression>, since it has no pairs to check. Assume $m\geq2$. If some $m$-term <arithmetic progression> has all its pairs blue, we are done. Otherwise, every such <arithmetic progression> contains a red pair. For each $(a,d)\in\mathbb N^2$, choose, say lexicographically, indices $0\leq r<s<m$ for which $\{a+rd,a+sd\}$ is red, and color $(a,d)$ by this chosen pair $(r,s)$. This is a <finite coloring> with at most $\binom m2$ colors.
We need a lattice pattern that works whichever pair becomes the common color. Fix, before applying the <Gallai theorem for an integer lattice>, the finite set
$$
F=\bigcup_{0\leq r<s<m}\{(si-rj,j-i):0\leq i,j<m\}\subseteq\mathbb Z^2.
$$
Translate $F$ into the positive quadrant and apply the <Gallai theorem for an integer lattice> to that translated pattern. Absorbing the translation into the base point gives integers $a_0,d_0$ and $q>0$ such that $(a_0,d_0)+qF\subseteq\mathbb N^2$ is <monochromatic>. Write $(r,s)$ for its common color. Since $(0,0)\in F$, the base point itself has positive coordinates.
For each $0\leq i,j<m$, the parameter point
$$
(a,d)=\bigl(a_0+q(si-rj),\ d_0+q(j-i)\bigr)
$$
belongs to that <monochromatic> copy and has chosen red pair $(r,s)$. Its red endpoints simplify to
$$
a+rd=a_0+rd_0+q(s-r)i,\qquad a+sd=a_0+sd_0+q(s-r)j.
$$
Consequently, putting
$$
A=\{a_0+rd_0+q(s-r)i:0\leq i<m\},\qquad B=\{a_0+sd_0+q(s-r)j:0\leq j<m\},
$$
gives two $m$-term <arithmetic progressions> of positive <common difference> $q(s-r)$, with every cross-pair red. These <arithmetic progressions> really are disjoint: the parameter point corresponding to $i=m-1,j=0$ has positive second coordinate, so $d_0>q(m-1)$. Hence
$$
\min B-\max A=(s-r)\bigl(d_0-q(m-1)\bigr)>0.
$$
This proves the \b[<arithmetic-progression bipartite Ramsey dichotomy>], including the required disjointness.
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