$a)$ Suppose $\left\{ m\left( a+\sqrt a \right) \right\}=\left\{ n\left( a-\sqrt a \right) \right\}$. This is equivalent to $\left\{ m\sqrt a \right\}=\left\{ -n\sqrt a \right\}$ so $m\sqrt a -(-n\sqrt a)=(m+n)\sqrt a\in \mathbb{Z}$ which is an obvious contradiction. $b)$ Suppose $\lfloor m\left( a+\sqrt a \right) \rfloor=\lfloor n\left( a-\sqrt a \right) \rfloor.$ Well let \[k=\lfloor m\left( a+\sqrt a \right) \rfloor=\lfloor n\left( a-\sqrt a \right) \rfloor.\]Using the properties of the floor function, we have \[k<m\left( a+\sqrt a \right)<k+1\text{ and }k<n\left( a-\sqrt a \right)<k+1\]which, dividing by $a+\sqrt a$ and $a-\sqrt a$ respectively and adding, we get \[\frac{k}{\frac{a-1}{2}}=\frac{2ak}{a^2-a}=\frac{k}{a+\sqrt a}+\frac{k}{a-\sqrt{a}}<m+n<\frac{k+1}{a+\sqrt a}+\frac{k+1}{a-\sqrt{a}}=\frac{2a(k+1)}{a^2-a}=\frac{k+1}{\frac{a-1}{2}}\]which is equivalent to \[k<\frac{a-1}{2}(m+n)<k+1\]which is a contradiction, because $a$ is odd, so $\frac{a-1}{2}(m+n)$ is an integer but so is $k$.

if m=n=0 aren't they true?

$m,m$ are at least one if they are natural. That's the old (pre 19th century) and correct definition of a natural number. This math exercise is translated from my mother tongue, Romanian, where they say "natural numbers minus zero", which is a wrong mindset. I'm not even saying "in my opinion".

Great problem! Counterexample for even $a$: $a=10$, $m=1$, $n=2$ (both sides are equal to $13$).