This is a question about pathologies.

Let $X/\mathbb{C}$ be an irreducible projective variety smooth over $\mathbb{C}$. Then, the singular cohomology groups $H^i(X, \mathbb{C})$ have a hodge decompositon, and hodge theory tells us that the betti numbers (and hodge numbers) are not completely random. Eg, the odd betti numbers $b_{2i+1}$ are even integers, Hard Lefschetz theorem, even betti numbers $b_{2i}$ are nonzero, ...

We can also see that the even Betti numbers of a variety $X$ as above are nonzero in "another" way: $X$ has a finite surjective map to $\mathbb{P}^n$ where $n= \dim X$. Then, because $X$ is Kahler, such a map induces an injective map on singular cohomology. Hence, the result follows from the corresponding result for projective space.

$\textbf{Question:}$

1.) Are there any nontrivial restrictions on the Betti numbers of smooth, irreducible proper (but non-projective) varieties?

For example, can I have such a 4-fold with Betti numbers $b_0 = b_8 = 1$, but all other $b_i = 0$?

I find the situation a little disconcerting: for example, suppose I have a connected compact topological (or complex) manifold. If I know it's betti numbers are as directly above, then I would like to immediately say that it doesn't have the structure of an algebraic variety. However, the only thing I can say now is that a smooth complex variety with these Betti numbers can't be projective. If such a variety existed, then (by Ehresmann's thm) we can't put it in a proper smooth family over a curve with any projective variety. I'm no expert but that sounds pretty bad.