Question: 1) Can a state be entangled without also being a superposition? (Please give an example.) 

2) Can a state be a superposition without being entangled? (Again, an example please.) 

3) And what about a cat state? 

I am royally confused, and a little bit of googling didn't help.  

This question is partly motivated by the recent news announcement (Nature, NYTimes) that ten billion entangled pairs have been created. 

4) Are these spin pairs also in a superposition? 

5) Does this count as a cat state?

Answer: 
This goes on order from the more general to the more specific:

1. If A and B are two states, then you can form any linear combination to get superposed state, let's call it C. Of course, any state C can be expressed as superposition of two other states A and B, so calling a state "superposed" is not all that meaningful, unless there is something distinguishing the A,B states from all their superpositions.
2. You can only discuss entanglement if you first distinguish two (or more) sub-systems of the full system, for example if the distinguished sub-systems do not interact with each other, or if they are spatially separated. In this case states of system (1) can be entangled with states of system (2), if the total state cannot be written as a product $A=A\_1 \times A\_2$ where $A\_1$ and $A\_2$ are states of the separate subsystems. A more formal way to quantify that is the "entanglement entropy", which tells you roughly how much information you are missing if you have access to one sub-system only.
3. The cat state is a particular example of an entangled state. Specifically in systems where two (or more) sub-systems consist of exactly one (qu)bit of information, which is labeled as 0 or 1, this is the state |00...0> +|11...1>. I am also told in comments there are generalizations beyond this system of n qubits, and more physical ways to characterize the state.

So, if you decide that your product states as above are distinguished over any linear combination thereof, there is a meaningful comparison between what you call "superposed" state (which I have to say is not a standard terminology) and the entangled states. An entangled state is one that cannot be expressed as product state, so it is necessarily a superposition of such states, say $A\_1 \times A\_2$ and $B\_1 \times B\_2$. The converse is not true: a superposition of product states may well be a product state itself.
The difference between a superposition and entanglement is the following. We consider a two slit experiment where a photon wave function interacts with a screen. The wave vector is of the form
$$
|\psi\rangle~=~e^{ikx}|1\rangle~+~e^{ik’x}|2\rangle
$$
as a superposition of states for the slits labeled $1$ and $2$. The normalization is assumed. The state vector is normalized as
$$
\langle\psi|\psi\rangle~=~1~=~\langle 1|1\rangle~+~\langle 2|2\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle
$$
The overlaps $\langle 1|2\rangle$ and $\langle 2|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate.

We now consider the classic situation where one tries to measure which slit the photon traverses. We have a device with detects the photon at one of the slit openings. We consider another superposed quantum state. This is a spin space that is 
$$
|\phi\rangle~=~\frac{1}{\sqrt {2}}(|+\rangle~+~|-\rangle).
$$
This photon quantum state becomes entangled with this spin state. So we have
$$
|\psi,\phi\rangle~=~e^{ikx}|1\rangle|+\rangle~+~e^{ik’x}|2\rangle|-\rangle
$$
which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state. Now consider the norm of this state vector
$$
\langle|\psi,\phi|\psi,\phi\rangle~=~\langle 1|1\rangle\langle +|+\rangle~+~\langle 2|2\rangle\langle-|-\rangle~+~ e^{i(k’~+~k)x}\langle 1|2\rangle\langle +|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2|1\rangle\langle-|+\rangle.
$$
The spin states $|+\rangle$ and $|-\rangle$ are orthogonal and thus $\langle +|-\rangle $ and $\langle-|+\rangle$ are zero. This means the overlap or interference terms are removed. In effect the superposition has been replaced by an entanglement.

This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state. So one does not need to invoke an outright collapse to illustrate how a superposition is lost. 

The cat state is a form of the GHZ state where you have
$$
|\psi\rangle~=~C(|0\rangle^{\otimes n}~+~|1\rangle^{\otimes n})
$$
which are n-partite entanglements. These types of states can violate Bell inequalities with one state, which illustrates quantum mechanics is not a purely statistical theory. The statistics are derived from QM.