Can an admissible SO(n) representation contain an SO(n-1) representation with infinite multiplicity? For simplicity, just let G be GL(n) over real numbers, K=SO(n),K'=SO(n-1). Now if $\pi$ is an admissible representation of G with respect to K, i.e., any irreducible K-representation occurs with finite multiplicity.
Now the question I want to know that: does any irreducible K'-representation occur with finite multiplicity?
 A: Not necessarily.  For example if $n = 2$, then $SO(1)$ is the trivial group,
and so if $\pi$ is infinite dimensional, it is the trivial representation of $SO(1)$ with infinite multiplicity.
Now suppose $n = 3$.  The trivial representation of $SO(2)$ occurs with mult. one in 
every irrep. of $SO(3)$.  So if $\pi$ is infinite dimensional, and so contains an infinite number of irreps. of $SO(3)$, 
then $\pi$ will contain the trivial representation of $SO(2)$ with infinite multiplicity.
A: Funnily enough, I wrote a paper, Cramped subgroups and generalized Harish-Chandra modules (published version:10.1090/S0002-9939-08-09421-5), about this question a few years ago.
The takeaway is that there is a geometric method for understanding when such a restriction is admissible.  For many pairs of groups it never is, but I think Matt found the only examples of the form SO(n) and SO(n-1).  In general, each finite dimensional representation of SO(n) comes from quantizing a coadjoint orbit, and you want only finitely many of the coadjoint orbits that lie in the image of the moment map on the contangent bundle of the sphere $T^*S^{n-1}$.  In particular, $T^*S^1 \to \mathfrak{so}_2^*$ is surjective since $S^1$ is a regular action, and $T^*S^2 \to \mathfrak{so}_3^*$ is surjective since the adjoint representation is covered by the orbit of any line.
