Question about transistors
The reason for the transistors is that the photo-resistors can't provide enough current to drive the motor. The transistors act as an amplifier of the small signal from the photo-resistors to make the signal large enough to power the motor directly.

These are darlington transistors, which will start to turn on when the voltage between the base (connected to the the photo-resistors) and the emitter (connected to the motor) reaches about 1.4V. They have very high gain, so only a very small current at the base is required to give a large current flowing from the collector to the emitter.

when both photo-resistors are at the same light level, the voltage between them is close to the middle the +5(VCC) and -5V(VEE) rails (zero volts). The emitters (connected together and connected to the motor) are also at zero volts, as they are connected through the motor to the ground (0V). In this state, there is no voltage between base and emitter to begin to turn on either of the transistors. 

When you cover one of the photo-resistors (say R2), the voltage at the junction of the two resistors moves towards the other rail (so if you cover R2, it's resistance increases, and the voltage at the junction  between R2 and R3 moves towards the -5V rail. 

Once this voltage moves below -1.4V, Q2 begins to turn on. Only a very small current at the base of Q2 is required to have the emitter of Q2 draw a large current from the collector. The transistor will draw as much current as needed from the collector to keep the voltage at the emitter 1.4V above the base, so as you reduce the light at R2, you lower the voltage at the base of Q2 which results in Q2 drawing more current from the -5V rail to keep the emitter at 1.4V above the base. This results in the motor turning faster.

Exactly the reverse happens  when you cover R3. Now the voltage at the base of Q3 (and Q2) increases, and Q3 begins to draw current from the +5V rail to turn the motor in the opposite direction.