Counting stationary and c.u.b sets I haven't seen this mentioned in Jech or Kunen, sorry if these are basic questions but I am not understanding these points.
If $\kappa$ is regular uncountable then in ZFC every stationary subset of $\kappa$ is the disjoint union of $\kappa$ stationary sets. So this holds for every successor $\gamma^+$. we have strictly less than $2^{\kappa}$ of them but is there a way to count how many stationary and c.u.b sets we can have? 
Is there a way of knowing which subsets of say $\omega_1$ are stationary? Do we know if a stationary set can contain a c.u.b set or the other way around or maybe it does not matter?
What about for a singular cardinal, for instance for $\aleph_{\omega}$, can we count the stationary subsets and the c.u.b sets? Would the GCH help in this case? 
 A: For every infinite cardinal $\kappa$, the number of club subsets of $\kappa$ is $2^\kappa$, fully as large as it could possibly be. Since every club set is stationary, this means also there are fully $2^\kappa$ many stationary sets also. 
To see that there are this many club sets, observe that there are $\kappa$ many successor ordinals below $\kappa$. For any set $A\subset\kappa$, let $A+1 = \{\alpha+1| \alpha\in A\}$' be the set of successors to elements of $A$. Let $C_A$ be the set consisting of $A+1$ together with all limit ordinals below $\kappa$. This is a club subset of $\kappa$, and it is easy to see that if $A\neq B$, then $C_A\neq C_B$. Since there are $2^\kappa$ many subsets $A$, it follows that there are $2^\kappa$ many club sets. This argument does not need GCH.
For stationary sets, you could have used Solovay's theorem directly. Since every stationary subset of $\kappa$ is a union of $\kappa$ many disjoint stationary sets, we can take any subunion corresponding to any subset of the index set, to get $2^\kappa$ many distinct stationary sets. 
About your latter considerations. Every club set is stationary; the intersection of two clubs is club and the intersection of a stationary and a club is stationary, and these are usually counted among the elementary facts about club and stationary sets. We usually do not consider the concept of stationary on singular cardinals of cofinality $\omega$, because in this case, the clubs do not form a filter. When $\kappa$ has uncountable cofinality, then the club filter makes sense, and the stationary concept is robust. The right way to think about it is: club means having measure $1$ with respect to the club filter, and stationary means not having measure $0$. Thus, stationary sets have outer measure $1$ with respect to club filter, although if they are co-stationary, then they will have inner measure $0$. 
One interesting issue is that whether or not a set $A$ is stationary can be affected by forcing. For example, for every stationary set $A\subset\omega_1$, there is an $\omega_1$-preserving forcing extension where it contains a club. Thus, if the complement of $A$ was stationary in the ground model, it becomes nonstationary in the extension. 
