I'm real suspicious that you should swap the capacitor and resistor that are connected to the collector and emitter of the middle transistor. (It sure would help if these components were numbered.) The base needs to be at a particular value to maintain the quiescent point, and I don't like the looks of the asymmetrical configuration. The way it's configured, it looks like the transistor on the left should be about 1 V below the transistor on the right. Off the top of my head, I don't see how that could be a good thing.
At what points have you probed the circuit?
juming said:
(you guessed it.. they've taught us nothing yet)

And they probably never will. Oh man! Have I been
there before! You're dragging up old unpleasent memories of late nights of studying separated by long days in the lab with an instructor that acts more like a magician than an electrician. It seems like you are so on your own, especially when they force you to work in those damn coalition groups.
juming said:
... we tried swapping the 1k resistor from the emitter of the second transistor to a 1k2 - I'm not sure what that was supposed to do though
That's called "tweaking." It's what all the great engineers do, don't you know.

Engineering design is a lot of guess-and-check. That is one thing they will most likely never come out and admit to their undergrad engineering students, at least, not officially. I wouldn't be able to tell you at this point if that was a "directed" guess or just an act of desparation. I would give your instructor the benefit of the doubt, though.
juming said:
... what's a curve tracer?
Your instructor should definitely be able to tell you that. If not, I would reconsider engineering schools. That is a rather important piece of design equipment for these rough discrete transistor topologies. Without one, you can basically forget about small signal circuit development (or shell out a lot of cash for highly toleranced components, in which case you might as well build an ASIC).
juming said:
... how would one calculate/find out the β for each transistor?
That's what you need to curve tracer for. You put the transistors on the curve tracer to find their quiescent points and the β's at those points. Then, you select your collector, emitter, and base components to maintain the quiescent point. Finally, your β value will determine the crossover coupling that you will get. I didn't actually happen to look at the parasitic capacitances of these BC337's, but I'm sure it's far less than 100 nF, so you probably don't have to worry about that. But you definitely need to know where you stand with quiescent point and β value.
Oh, and even though you didn't ask, I'll go ahead and tell you anyway. β is the small signal gain from base to collector. Actually, it's been so long that I can't remember if that's current gain, voltage gain, transimpedance, or transconductance. I'm pretty sure it's voltage gain (the small signal component, so subtract the quiescent bias).
juming said:
... the difference between nominal & actual values of components?
This is one of those things that distinguishes good engineers from the rest. As an engineer, one of your fundamental distinctions is 1) the realization that nothing is what it says it is or should be, and 2) you know how to make it work regardless of this problem and keep it transparent to your customer. Once you graduate, this is one of those distinctions that you will (hopefully) get paid for. It is quite trivial; the real trick is #2. As an example, there are 1 kΩ pull-ups. They may actually be 986 Ω and 1003 Ω. This is probably critical to your circuit since it relies on the crosscoupling of the collector being pulled up by a 986 Ω resistor and the collector being pulled up by a 1003 Ω resistor. On first order, this will negatively affect the quiescent point (which is semi-critical). But, on second order, it will affect the swing of the collector voltage, which will get amplified at the collector of the other transistor through its gain. I think you really need to get ahold of a curve tracer for this one, but I may just be paranoid. It has been a while, like I said.