Resistor value calculation
The transistors in the circuit are in a Darlington configuation, which means that the signal that goes into the first amplifier is greatly amplified.

The form of transistor biasing that is used here seems to be fixed bias, or base bias.

Here are some equations that would apply here (for each transistor):

\$I_C = \beta I_B\$

\$V_{CE} = V_{CC} − I_C R_C\$

The first transistors must amplify a very, very weak current. The beta (\$\beta\$) for these transistors probably ranges from 75 to 200, which means that the large resistor allows the the transistor to turn on. If the resistor was tiny or non-existent, it would change the bias of the transistor and cause it to have a hard time turning on.

Also, the resistors help limit current. In the event that there was no current limit the first transistor is turned on, that would possibly destroy the first two transistors because a lot of charge will flow from \$V_{CC}\$ (6 volts) to ground.