OK, using the original battery pack - w/o resistors.
What is the voltage across the LED string (or across the battery) under load? Still about 3V?
I didn't check this, I only checked the current using the clamp.
Why AC and not DC? (particulary since it is connected to the battery pack)
That is he dial setting for this model meter... it also has the DC sign on it, but I don't know how to type it you just press the blue select button to change to DC.

Ok, I have worked out some [possibly very incorrect, and stupid] numbers!

Please don't shoot me, I am still learning

I didn't even know what a resistor was/did 4 days ago, so bare with me

Let's start with the individual calculations of running each circuit in
parallel.Line 1 - 10 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 41mA read from Uni-T clampResistor needed: 6V-3V = 0.041 x R ==> R = 3V / 0.041 = 73 Ohms
Wattage of resistor needed: 0.041 x 0.041 x 73 = 0.122W - so an R82 (or R68?) 0.5W should be correct for this?
Line 2 - 10 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 36mA read from Uni-T clampResistor needed: 6V-3V = 0.036 x R ==> R = 3V / 0.036 = 83 Ohms
Wattage of resistor needed: 0.036 x 0.036 x 83 = 0.107W - so an R82? 0.5W should be correct for this?
Line 3 - 6 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 37mA read from Uni-T clampResistor needed: 6V-3V = 0.037 x R ==> R = 3V / 0.037 = 81 Ohms
Wattage of resistor needed: 0.037 x 0.037 x 81 = 0.110W - so an R82 0.5W should be correct for this?
Line 4 - 15 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 41mA read from Uni-T clampResistor needed: 6V-3V = 0.041 x R ==> R = 3V / 0.041 = 73 Ohm
Wattage of resistor needed: 0.041 x 0.041 x 73 = 0.122W - so an R82 (or R68?) 0.5W should be correct for this?
BUT: We want 1 + 2 in Series and 3 and 4 in Series:1 + 2 = Circuit A
3 + 4 = Circuit B
Circuit A Total consumption = 77mA
Resistor needed: 6V - 3V = 0.077 X R ==> R =3V / 0.077 = 38.96 Ohm (So 39 Ohm)
Wattage of resistor needed: 0.077 x 0.077 x 39 = 0.231W - So an R39 0.5W should be sufficient for Circuit A
Circuit B Total consumption = 78mA
Resistor needed: 6V - 3V = 0.078 X R ==> R =3V / 0.078 = 38.46 Ohm (So 39 Ohm)
Wattage of resistor needed: 0.078 x 0.078 x 39 = 0.237W - So an R39 0.5W should be sufficient for Circuit B
So, going back to my rubbish diagram, I need to add 2 x R39 0.5W resistors where the green dots are, and they shouldn't get hot because they aren't under much load?

Am I correct in thinking that we don't need to limit the PSU? Each individual line of LEDs is run in series, so a failure would take out the whole line? Would then not the entire 6V be dumped on the remaining 2 lines.... or am I getting this totally wrong?
Edit: Oops, I typoed one of the resistor values as R9 instead of R39, fixed.