Solution:
Answers: $[N/2]$, $[(N+1)/2]$, $[N/2]$.

Suppose $A$ leaves piles $n$, $m$ with $n \leq m$.

Under $R1$, $B$ can certainly secure $m$ by dividing the larger pile into $1$ and $m-1$. He cannot do better, because if $b$ is the biggest of the $4$ piles, then the smallest is at most $m-b$. Hence $A$'s best strategy is to leave $[N/2]$, $[(N+1)/2]$.

Under $R2$, if $A$ leaves $a=2$, $b=N-2$, then $B$ cannot do better than $[N/2]$, because if he divides the larger pile into $a$, $b$ with $a \leq b$, then he takes $a+1$. $A$ cannot do better, because if he leaves $a$, $b$ with $3 \leq a \leq b$, then $B$ can divide to leave $1$, $a-1$, $[b/2]$, $[(b+1)/2]$. Now if $a-1 \geq [(b+1)/2]$, then $B$ takes $b \geq [(N+1)/2]$. If $a-1 < [(b+1)/2]$, then $B$ takes $a-1 + [b/2]$. But $a-1 \geq 2$ and $[b/2] \geq [(b+1)/2] - 1$, so $a-1 + [b/2] \geq 1 + [(b+1)/2]$, or $B$ takes at least as many as $A$, so $B$ takes at least $[(N+1)/2]$.

Under $R3$, $A$'s best strategy is to divide into $[N/2]$, $[(N+1)/2]$. We have already shown that $B$ can secure $[(N+1)/2]$ and no more by following $R1$. He cannot do better under $R2$, for if he divides so that the biggest pile comes from $[N/2]$, then the smallest does too and so he gets $[(N+1)/2]$. If he divides so that the biggest and smallest piles come from $[(N+1)/2]$, then he gets only $[N/2]$. But one of these must apply, because if he divided so that the smaller from $[N/2]$ was smaller than the smaller from $[(N+1)/2]$, and the bigger from $[N/2]$ was smaller than the bigger from $[(N+1)/2]$, then $[N/2]$ would be at least $2$ less than $[(N+1)/2]$ (which it is not).