1) I think that the 100u capacitor is purely there to smooth the input voltage. If that’s the case then I don’t think I want that.
Yep. In your case you may not need it.
2) I think the 1u capacitor is only needed to create the flash. I’m happy to keep that.
It essentially forms an RC circuit with one of the resistors and determines the speed of the flashing.
3) A simple potential divider defines the activation point of the first (leftmost) transistor. Why is this 4V not 4.6V?
4.6*150k/(150k+1M)=0.6V at the transistor base.
Build it and try it. Your calculation seems reasonable. 4*150/1000 = 0.6. That may be an error on the part of the designer.
4) When the 1st transistor is closed (C-E connected), I think that the current into the 2nd base is zero as all of the flow passes through the 1st transistor.
Don't think of transistors as open or closed. They allow more or less current through them depending on the base current and their gain. Essentially the collector current is the base current multiplied by the gain of the transistor. When more current *could* flow than the rest of the circuit allows, the transistor is effectively saturated.
5) When the 1st transistor opens, the current then passes into the 1u capacitor (still nothing into the 2nd base as yet) until the capacitor is charged.
Call the transistors Q1, Q2, and Q3 from left to right.
When the voltage is above 4.6V, Q1 is turned on, robbing base current from Q2, so Q2 remains off. With Q2 off, there is no base current to Q3, so it also remains off.
Because Q3 is off, the +ve end of C1 is effectively connected to ground via the 33ohm resistor. As soon as Q1 turns off, C1 begins to charge through the 330K resistor and the voltage at the base of Q2 rises toward 0.6V. As it does so, Q2 begins to turn on, providing current to Q3 which also begins to turn on.
As Q3 begins to turn on, the additional current across the 33R resistor pulls the +ve end of the capacitor up. The other end (which is actually more positive) then begins to discharge into the base of Q2 (which now sees a higher voltage) and gets turned on harder. This increases the base drive to Q3 which increases the current through the 33R resistor which causes more current to be available on the base of Q2... So Q2 and Q3 get turned on hard.
However, the charge in C1 is not limitless. As it begins to discharge, eventually the current available to the base of Q2 falls, which causes it to turn off a little. This turns off Q3 a little and the voltage across the 33R resistor falls a little. (By this time the +ve end of C1 is more positive) As the voltage across the 33R resistor falls, the voltage on the other end of C1 falls/ This robs more current from the base of Q2 (it gets diverted to C1) which causes Q2 to turn off some more, and Q3 to turn off some more, which causes the voltage across the 33R resistor to fall some more... Both transistors get turned off hard.
Then the 330K resistor starts charging C1 again, and the cycle starts again...
7) The current through the LED will be 4V (or 4.6V?) /33 = 121mA
Presuming the LED drops 2V, the max current is 2.6/33 = 78mA. It will be less than that because there is some voltage drop across Q3. This represents the peak current, but the LED is more often off than on, so the average current is much lower. Incidentally, this is the reason for the 100u capacitor. It is there to provide the current that the almost flat batteries could not. In your case this current is probably small compared to the load current so you could omit it.
8) I don’t know what the 1k resistor is for?
It is to limit the base current to Q3
Sorry for another load of questions but I think I understand part of this, just not enough that I’m happy to continue with another) circuit design. At a guess I’d say that I could lose the two capacitors, increase the 33Ohm resistor and reduce the 150k resistor and end up with a functioning (but not flashing) circuit. Unfortunately I couldn’t say for sure that that would work and I’m also interested to understand exactly how the posted circuit does work.
Keep the 1u capacitor. Increase the 33R resistor to limit the max LED current. You do have to consider the maximum reverse voltage applied to the BE junction of Q2. Ensure that your low voltage limit is no higher than a volt or two higher than the maximum reverse voltage allowed for this junction (it's often around 6 to 8 V, and it's 6V for a BC548).
Using a blue LED would be a good idea. Essentially the max trigger voltage should be less than 6V + LED voltage.
For now I'll keep looking through circuits and reading up on transistors but if you could mark my answers as it were, that would be a great help!
I hope that helps.