= Solution
The integral <Poincare duality> theorem says that a closed connected oriented $n$-manifold has a <fundamental class> $[M]\in H_n(M;\mathbb Z)$ for which cap product gives $H^q(M;\mathbb Z)\cong H_{n-q}(M;\mathbb Z)$. The general <Poincare duality with the orientation local system> uses $[M]\in H_n(M;\mathcal O_M)$ and gives
$$
H^q(M;\mathcal A)\xrightarrow{\ \frown[M]\ }H_{n-q}(M;\mathcal O_M\otimes\mathcal A)
$$
for a <local coefficient system> $\mathcal A$. Over $\mathbb F_2$ it holds without orientability. For constant integral coefficients on a nonorientable connected closed manifold, $H_n=H^0(M;\mathcal O_M)=0$, while $H^n=H_0(M;\mathcal O_M)=\mathbb Z/2$: the invariants of sign reversal in $\mathbb Z$ are zero and its coinvariants are $\mathbb Z/(a\sim-a)$.
If the three-manifold is orientable, duality and the <universal coefficient theorem for cohomology> give
$$
H_2(M;\mathbb Z)\cong H^1(M;\mathbb Z)\cong\operatorname{Hom}(H_1(M;\mathbb Z),\mathbb Z)\cong\boxed{\mathbb Z^b}.
$$
For a nonorientable three-manifold, $H_3=0$ and $H^3=\mathbb Z/2$. The top-degree universal coefficient sequence therefore gives
$$
\operatorname{Ext}^1_{\mathbb Z}(H_2(M;\mathbb Z),\mathbb Z)\cong\mathbb Z/2.
$$
Since the homology groups are finitely generated, their cyclic decomposition shows that this Ext group is isomorphic to the torsion subgroup of $H_2$. Thus the torsion is exactly $\mathbb Z/2$, not the possibly larger torsion subgroup of $H_1$. Also <odd-dimensional closed manifolds have zero Euler characteristic>, by pairing complementary mod-two Betti numbers. If $r$ is the free rank of $H_2$, it follows that $0=\chi(M)=1-b+r$, giving the <second homology of a closed nonorientable three-manifold>
$$
\boxed{H_2(M;\mathbb Z)\cong\mathbb Z^{b-1}\oplus\mathbb Z/2,\qquad b\geq1.}
$$
The inequality is forced by the nonnegative free rank $b-1$.
For the final graded-group construction, one always has $K^0=K^3=K^6=\mathbb Z$, since each of these is copied from $H^0$ of a connected three-manifold. Suppose it were the cohomology of a closed six-manifold $N$. The group $K^0$ makes $N$ connected, and $K^6=\mathbb Z$ makes it orientable: the nonorientable alternative has top integral cohomology $\mathbb Z/2$, as above. Its third rational cohomology would then be one-dimensional. But the middle <Poincare duality pairing> on $H^3(N;\mathbb Q)$ is nondegenerate and skew-symmetric by graded commutativity, hence must have even dimension. Equivalently, a generator $u$ would satisfy $u\smile u=-u\smile u$, so its square in the torsion-free top group is zero and the entire rank-one pairing vanishes. This contradicts duality and the <even rank of middle cohomology in dimension four k plus two>. Therefore
$$
\boxed{K\text{ is never the integral cohomology of a closed six-manifold}.}
$$
Both existence questions have negative answers, independently of the other groups of the original three-manifold.
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