1) You can pick the homology below the middle dimension quite arbitrarily. More precisely, given a finite complex $K$ and a number $n$, there exists a closed, parallelizable $2n$-dimensional manifold $M$ and an $n$-connected map $f:M \to K$. You begin with a constant map $S^{2n} \to K$ and make it more and more connected by surgeries.
2) as you said, a necessary condition for the homology of a manifold is Poincare duality. If you have a finite complex $X$ that satisfies Poincare duality, the question of whether there is a smooth manifold homotopy equivalent to $X$ is a basic problem in surgery theory. If $X$ is simply connected, this has largely been solved by Browder. The answer is that if $X$ is odd-dimensional, there is such a manifold; and if the dimension is divisible by $4$, there is a manifold precisely if there is a stable vector bundle on $X$ such that the Hirzebruch signature formula holds with this bundle. In dimensions $2,6,10,\ldots$, there is a subtle problem with the "Kervaire invariant". And: I forgot to say that the dimension has to be at least $5$. For nonsimplyconnected complexes, Wall gave at least a theoretical answer.
3) Poincare duality for integral coefficients (and closed oriented $M$) says that $H_i (M) \cong H^{n-i}(M)$. The universal coefficient theorem implies that the torsion subgroups (for each space with finitely generated homology) are $T H^{i+1} = T H_i$ (abstract isomorphism). Combined, these two results tie the torsion subgroups of cohomology together.
4) I would not say that compactly supported cohomology contains more information than ordinary cohomology - they contain different information. With rational coefficients, you have an isomorphism $H^i(M) \cong (H^{n-i}{c}(M))^{\ast}$; H^{n-i}_{c}(M))^{\ast}$; the other isomorphism$H^{i}{c}(M) H^{i}_{c}(M) \cong (H^{n-i}(M))^{\ast}$holds iff the cohomology vector space are finitely generated. 1 1) You can pick the homology below the middle dimension quite arbitrarily. More precisely, given a finite complex$K$and a number$n$, there exists a closed, parallelizable$2n$-dimensional manifold$M$and an$n$-connected map$f:M \to K$. You begin with a constant map$S^{2n} \to K$and make it more and more connected by surgeries. 2) as you said, a necessary condition for the homology of a manifold is Poincare duality. If you have a finite complex$X$that satisfies Poincare duality, the question of whether there is a smooth manifold homotopy equivalent to$X$is a basic problem in surgery theory. If$X$is simply connected, this has largely been solved by Browder. The answer is that if$X$is odd-dimensional, there is such a manifold; and if the dimension is divisible by$4$, there is a manifold precisely if there is a stable vector bundle on$X$such that the Hirzebruch signature formula holds with this bundle. In dimensions$2,6,10,\ldots$, there is a subtle problem with the "Kervaire invariant". And: I forgot to say that the dimension has to be at least$5$. For nonsimplyconnected complexes, Wall gave at least a theoretical answer. 3) Poincare duality for integral coefficients (and closed oriented$M$) says that$H_i (M) \cong H^{n-i}(M)$. The universal coefficient theorem implies that the torsion subgroups (for each space with finitely generated homology) are$T H^{i+1} = T H_i$(abstract isomorphism). Combined, these two results tie the torsion subgroups of cohomology together. 4) I would not say that compactly supported cohomology contains more information than ordinary cohomology - they contain different information. With rational coefficients, you have an isomorphism$H^i(M) \cong (H^{n-i}{c}(M))^{\ast}$; the other isomorphism$H^{i}{c}(M) \cong (H^{n-i}(M))^{\ast}\$ holds iff the cohomology vector space are finitely generated.