Assuming the Axiom of Choice, every cardinal is either finite (i.e., an element of $\omega$) or Dedekind-infinite (i.e., in bijection with a proper subset of itself). This dichotomy is not true in ZF, however; in particular, it is consistent with ZF that there exist sets which are non-finite but cannot be partitioned into two non-finite pieces. Such sets are called "amorphous," and models of ZF containing amorphous sets can be constructed by building a permutation model of ZFA (ZF with urelements) and applying the Jech-Sochor embedding theorem.
Naturally, there are lots of types of structure which amorphous sets simply cannot have. For example, no amorphous set can be linearly ordered, since a linear order on a non-finite set allows us to either inject $\omega$ into the set, or partition the set into a "left" and "right" piece, both of which are non-finite. However, amorphous sets can still have some structure. For example, say that a set is "even" if it can be written as a disjoint union of 2-element subsets. Then, again using permutation models, we can construct a model of ZF which contains an even amorphous set (the lack of choice prevents us from using the evenness of the set to partition it into two non-finite pieces).
My question is twofold. First, say that a set is "odd" if it can be written as the disjoint union of a singleton and an even set.
Question 1: Is ZF+"there exists an amorphous set"+"every set is either even or odd" consistent? (It occurs to me that this may depend on whether the above "or" is meant inclusively or exclusively.)
My main issue with this question is that I don't see how such a model would be constructed. For example, we can use a permutation model to create a single amorphous even (or odd) set, but how do we then conclude that every set is either even or odd? I don't know how to do this at all, so this motivates my second question:
Question 2: How does one build models of ZF+"there exists an amorphous set"+"every set is ---," in general? (Where --- denotes some arbitrary niceness property.)