First, angular momentum isn't measured about an axis. It's measured about a point.
Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in.
Now, about your example. The total angular momentum, on the three block system, is definitely conserved. Why wouldn't it be? What you're perhaps missing is the fact that there will be in general an angular momentum associated with the 3rd block as well, even if it's traveling on a straight line. For example, say you have a block with momentum $\vec{p}$ traveling in a straight line at constant velocity until it hits the block at one end of the string (at the point of closest approach to the center of mass). Say the string has length $2\ell$. Now let's calculate the angular momentum of this third block about the center of mass of the two rotating blocks when it's at a distance $r$ from that center of mass.
$$\vec{L} = \vec{r} \times \vec{p} = rp \sin \theta \hat{z} = rp \frac{\ell}{r}\hat{z} = rp \hat{z} $$
(The $\hat{z}$ is just my choice in how to orient the system in space).
Notice now that $L$ does not depend on the distance $\vec{r}$. It's a constant of the motion, as promised, so long as $\vec{p}$ itself is conserved. Notice that there was nothing special in this derivation about the center of mass: it could've been literally any point and the conclusion still would be valid.
Now you can work out what will happen in the collision, assuming that linear momentum is conserved, and you'll prove explicitly that angular momentum is conserved in this process.
For (b), obtain the velocity of the CM by momentum conservation. Then just use your answer in (a) to find the velocities relative to the CM, and then add the velocity of the CM, you get the answer of (b). Yes, it is instantaneously rotating about $A$.
For (c), it's similar to (b). The velocity is the vector sum of the CM velocity you used in (b) and the velocity relative to CM due to rotation about CM.
Best Answer
Angular momentum should be conserved in any inertial frame of reference - if you move with the center of mass, the motion you see will be rotation about the center of mass; if you move with a different frame of reference, you will see rotation about a different axis.
So the short answer is "it doesn't matter".
Whether they "give the same answer" depends on whether you do the calculations properly...