The answer to 1 is "yes", and 2 is not quite well-defined, to my mind.

I'm going to follow Takesaki, Chapter III, Section 4, but this is all standard stuff. Given a C*-algebra $A$, a closed subspace $V$ of $A^*$ is *left-invariant* if $a\mu\in V$ for each $a\in A,\mu\in V$, where $(a\mu)(b) = \mu(ba)$ for $b\in A$. Then we have the following (Corollary~4.4 in the book):

**Claim:** $V$ is the closed left-invariant subspace generated by $V \cap A^*_+$.

In particular, $V$ certainly contains states. If $I\subseteq A$ is a *right* ideal then the annihilator $I^\perp = \{\mu\in A^* : \mu(a)=0 \ (a\in I) \}$ is left-invariant, and so

**Claim:** For each proper closed right ideal $I\subseteq A$ there is a state $\mu\in A^*$ which annihilates $I$.

So, if $f:A\rightarrow B$ is a degenerate $*$-homomorphism, then let $I$ be the closed linear span of $f(A)B$. By assumption $I$ is not equal to $B$, and obviously $I$ is a right ideal. So, there is some state $\mu\in B^*$ annihilating $I$, in particular, with $\mu\circ f=0$.

(If $B$ is non-unital, I guess I need to argue using approximate identities to show that $f(A) \subseteq I$. Also, using Cohen-Hewitt factorisation it follows that $I = \{ f(a) b : a\in A, b \in B \}$, no linear span or closure needed.)

So, for question 2. As every member of $A^*$ is the linear combination of (at most) 4 states, $S(A)$ certainly separates *points* of $A$. But if by "positive linear functionals separate points from closed subspaces" we mean that we want our functional to annihilate the subspace, then it's very easy to come up with counter-examples.