Is total boundedness a topological property?

The answer to your first question is "no". The real numbers are homeomorphic to the open interval $(-\frac{\pi}{2},\frac{\pi}{2})$, via the bijection given by $x\mapsto \arctan(x)$. Both spaces have a metrizable topology. But $(-\frac{\pi}{2},\frac{\pi}{2})$ is totally bounded, whereas $\mathbb{R}$ is not. Equivalently, endow $\mathbb{R}$ with the metric $d(a,b) = |\arctan(a)-\arctan(b)|$; this metric induces the same topology on $\mathbb{R}$ as the usual metric, but whereas $\mathbb{R}$ is bounded under $d$, it is not bounded under its usual metric.

Second question was answered inter alia elsewhere.

Consider the positive integers with the discrete topology. On the one hand this topology is induced by the usual metric, which is certainly not totally bounded. On the other hand, it is also induced by the metric $$d(n,m)=\left|\frac1n-\frac1m\right|\;,$$ which clearly is totally bounded.

This is really just the observation that $\left\{\dfrac1n:n\in\mathbb{Z}^+\right\}$ and $\mathbb{Z}^+$ are homeomorphic as subspaces of $\mathbb{R}$ with the Euclidean topology, via the map $n\mapsto\dfrac1n$, where the first subspace has compact closure in $\mathbb{R}$.

Others have pointed out that total boundedness is not a topological property; what's also worth noting is that it is a uniform property. A uniform space is a space that carries some additional structure beyond its topology that allows you to make sense of such notions as uniform continuity, uniform convergence, total boundedness, Cauchy sequences/nets/filters, completeness, completion...

I'll leave the precise definition to the linked article (there are several equivalent formulations), but the point is that a metric space (more generally, pseudometric space) naturally carries the structure of a uniform space (it's kind of the obvious motivating example -- the other motivating example being topological groups), and that total boundedness is a uniform property. (Worth noting is that so is completeness, and the theorem that a metric space is compact iff it's complete and totally bounded works also for more general uniform spaces. So do a lot of the basic metric space theorems, really, though unfortunately not all of them.)

(By contrast, boundedness isn't even a uniform property, since every metric space, bounded or not, is uniformly homeomorphic to a bounded metric space.)