Let's write $M = H_3(\mathbb{R})/H_3(\mathbb{Z})$. The manifold $H_3(\mathbb{R})$ is a Lie group, so it is certainly orientable. (An $n$-manifold is orientable if the structure group of its tangent bundle can be reduced to $\operatorname{GL}_n^+(\mathbb{R})$. But a Lie group has a trivial tangent bundle: the structure group can be reduced to $\{1\}$!) The elements of $H_3(\mathbb{Z})$ are translations on the Lie group $H_3(\mathbb{R})$, so each one is an orientation-preserving diffeomorphism. It follows that $M$ is orientable. $M$ is also clearly compact and connected, so Poincare Duality applies:
$H_0(M,\mathbb{Z}) \cong H_3(M,\mathbb{Z}) \cong \mathbb{Z}$,
$H_1(M,\mathbb{Z}) \cong H_2(M,\mathbb{Z}) \cong H_3(\mathbb{Z})^{\operatorname{ab}}$.
So it remains to compute the abelianization $H_3(\mathbb{Z})^{\operatorname{ab}}$. If you look at the generators and relations for $H_3(\mathbb{Z})$ given in the wikipedia article, it is almost immediate that the abelianization is $\mathbb{Z}^2$.
Also, there are many important classes of 3-manifold; does the Heisenberg manifold belong to some of them?
Well yes, of course. That's a terribly vague question, and I'm not an expert here, but the Heisenberg group is the simplest example of a compact nilmanifold beyond tori. Nilmanifolds are important in a wide variety of areas of mathematics, e.g. in group cohomology, geometry and algebraic topology. Remarkably, in the last several years they have been studied by Green and Tao for their applications in additive combinatorics! The aforelinked wikipedia article contains some references, enough for you to know much more about this subject area than I do.