Another way of reducing/limiting the solenoid current, albeit more complicated than just a resistor, is PWM (pulse width modulation). This has the advantage of dissipating less power than simply a resistor.
It looks correct, apart from the fact that there should be a reverse-biased diode (e.g a 1N4148) across the relay coil to protect the BC547 (and possibly the IC) from the back-emf voltage spike created when the coil is de-energised . Good luck.
That is about the first statement you have made which is correct ! :)
But if you want other people to show an interest in your circuit you should make it easy for them by drawing the circuit in a conventional way which can be readily understood without the need for mental gymnastics.
Welcome to maker.pro.
As Crutschow says, start with basic electronics, such as learning the function of individual components.
Chargers for laptops and phones are actually inside those devices and just need the appropriate power supply. You don't build the chargers themselves, and they are...
Without the flywheel diode (as per Danadak, post #3) the voltage spike when the relay coil switches off could have damaged the IC.
You might also need a decoupling capacitor (several uF) connected directly between pins 8 and 16)
No. Just remove/disconnect the diode, so that the chanter cap voltage isn't fed to the mixer. Don't bypass the diode. You don't want the drone circuit to affect the chanter cap voltage.
I think that may be the source of your tremolo effect. The drone circuit will, via the mixer, source or sink some current from the chanter timing cap and hence modulate the chanter signal. Try disconnecting the diode from the timing cap and see if the tremolo disappears.
If you are using the MM on a resistance range, don't forget the meter itself includes a current source to pass current through the load under test. Also, are the loads floating, or ground-referenced?