I got this schematic from the web and I want to know how it works. I included calculations I've done in the attachments.

I want to know what mode of operation the transistors are in. I'm assuming they're in forward active, but I don't think there is any current going through the collector for either of the transistors because I don't think there is no current going through the base. So, I'm a bit confused.

I also want to know how this schematic is better than using a voltage divider to get a dual rail power supply.

 

I got this schematic from the web and I want to know how it works. I included calculations I've done in the attachments.

I want to know what mode of operation the transistors are in. I'm assuming they're in forward active, but I don't think there is any current going through the collector for either of the transistors because I don't think there is no current going through the base. So, I'm a bit confused.

I also want to know how this schematic is better than using a voltage divider to get a dual rail power supply.

You've got the right idea, but in the real world, diode Vf and transistor Vbe aren't necessarily exactly a 0.7 volt drop. Diode drop may be higher than that and the transistor drop may be lower. The 1 k resistors will compensate for the difference. So, the transistors may be biased on just a bit.

The transistor circuit may be a bit better than the resistor only divider.
A better rail splitter is one which uses an op-amp, because the op-amp circuit uses feed back to compensate for changes in current draw.

 

Here's an op-amp rail splitter:
http://sound.westhost.com/project43.htm

 

I also want to know how this schematic is better than using a voltage divider to get a dual rail power supply.

The voltage divider allows only a little current to flow through the middle potential. Most of it is through -Vcc still. The first circuit will allow more current through the middle.

 

The transistor circuit shown would not be very 'stiff' - it's output impedance is about 500 ohms, any appreciable load will cause the output voltage to deviate.

You are on the right track, you basically want to create a voltage follower with a low output impedance that can source and sink current. The circuit also needs to be stable when driving a large capacitive loads, as you will inevitably have bypass capacitance in the powered circuit.