I start by observing $\lfloor (4+2\sqrt{3})m \rfloor = \lfloor(4-2\sqrt{3})n \rfloor$ iff there exists a $k\in\mathbb{N}$ such that $$ k<(4+2\sqrt{3})m<k+1\quad\text{and}\quad k<(4-2\sqrt{3})n<k+1. $$This is true if and only if, $$ \frac{k(2-\sqrt{3})}{2}<m<\frac{(k+1)(2-\sqrt{3})}{2} \quad\text{and}\quad \frac{k(2+\sqrt{3})}{2}<n<\frac{(k+1)(2+\sqrt{3})}{2}. $$We now attack the problem, starting from ${\rm (b)}$: ${\rm (b)}$: Note that $2k<m+n<2k+2$. From here, $m+n=2k+1$, is clearly odd. ${\rm (a)}$: Let $I_k = (k(2-\sqrt{3})/2,(k+1)(2-\sqrt{3})/2)$ and $I_k'=(k(2+\sqrt{3})/2,(k+1)(2+\sqrt{3})/2)$. It suffices to establish there exists infinitely many positive integers $k$ for which $|I_k\cap \mathbb{N}|\geqslant 1$ and $|I_k'\cap \mathbb{N}|\geqslant 1$. Note that the length of $I_k'$ is $1+\sqrt{3}/2$, larger than one, hence, $I_k'$ always contains an integer. We now turn our attention to $I_k$. Note that $I_k\cap \mathbb{N}\neq \varnothing$ iff $$ \left\lfloor \frac{k(2-\sqrt{3})}{2}\right\rfloor +1 \in I_k \iff \{k\alpha\}>2/\sqrt{3}, $$where $\{t\}=t-\lfloor t\rfloor$ is the fractional part of $t$, and $\alpha=(2-\sqrt{3})/2$. But this is clear: it is well-known that the sequence $\{n\alpha\}_{n\geqslant 1}$ is dense in $(0,1)$ provided $\alpha$ is irrational (the proof of this fact is elementary through pigeonhole principle -- DIY!).

For a) : Let $f( n) =\left( 4-2\sqrt{3}\right) n$. Then we have $f( n+1) -f( n) \leq 1$, $f( 1) < 1$, and $\lim _{n\rightarrow \infty } f( n) =\infty $. Therefore, for any $r\in \mathbb{N}_{\geq 0}$, we can take $n\in \mathbb{N}$ such that $r=\lfloor f( n)\rfloor $. $\blacksquare$