Can every uncountable cardinal be realized as a cofinality? I searched a bit, but couldn't find an answer to the question in the title.
Given an uncountable cardinal $\kappa$ is there an ordered set with cofinality equal to $\kappa$?
 A: Sometimes. There is an extra condition required for an uncountable cardinal to appear as a cofinality (that is, the least order-type of a unbounded subset.) Namely the cardinal must be regular. 
Definition: A cardinal $\lambda$ is regular, provided, every subset with order-type strictly less than $\lambda$ is bounded.
Proposition: If $(L, <)$ is a totally ordered set, then the cofinality of $L$ is regular. 
Proof: Let $\lambda$ be the cofinality of $L$; moreover assume $C=\{a_\alpha: \alpha \in\lambda \} \subset L$ is cofinal with order-type $\lambda$. Then, every unbounded subset $X\subset \lambda$ of $\lambda$ generates a new cofinal subset of $L$, namely $C_X = \{ a_\alpha: \alpha\in X\} \subset C$. Moreover, this new cofinal subset, $C_X$, has the same order-type as $X$; by the minimality of $\lambda$, each cofinal subset $C_X$ must again have order-type $\lambda$, establishing the regularity of $\lambda$.
Edit:
I just realized that you didn't specify that the order needed to be total. If you remove that requirement, then yes every uncountable cardinal can occur as the cofinality of an ordered-set.
For example: 
The set of finite decreasing sequences of ordinals less than $\aleph_\omega$, ordered by end-extension has cofinality $\aleph_\omega$ (here the cofinality reverts back to the least cardinality of a cofinal subset.)
