Electronic – 3.3 Volts Output on 1.8 Volt Regultor

1. Do not use a voltage divider circuit to power something like an RF transmitter, that will not work during pulsed currents as the "load" equivalent resistance drops when it needs more current, and therefore the divider will not actually "Regulate" at all.

2. You can do this, if you do not trust your 7805's other parallel loads, and this will probably help protect the 5V system from noise feedback from the RF system, which you may get in the #3 option. In this method, make sure both regulators have good input capacitance, and also reasonable output capacitance (follow datasheet recommendations, usually put a little extra to avoid spike brownouts which often happen, especially with RF modules).

3. yes, I would put the LDO after the 5V 7805's output, as long as the dropout is actually within spec. Go for one with less than 1V drop out and you should be fine. Ensure plenty of capacitance between the two regulators, and on the LDO's output stage to feed the pulsed TX currents of the NRF24.

These 7V wall warts I assume have enough output current to run your system. You just mention some 5V PIC-based circuits, which I assume are low-powered. The NRF24 is low power, but has high power bursts which can brown-out and cause things to reset (just see all the other questions on EE Stack Exchange about power issues with RF TX/RX modules hehe). To fix this, always ensure good and fast acting capacitors are available nearby, and are properly rated.

Capacitor chemistries like tantalum and ceramics are the fastest to respond to pulse currents, and have the least ESR (equivalent series resistance, this is effectively what slows down other capacitor types). A cheapo electrolytic capacitor has very high capacitance but often poor ESR characteristics, making them slow to react and therefore their "reactance" at high frequency (fast pulsed loads) means they are almost useless.

Your 3.3V system should be okay from either the 7V wall wart output or the 5V 7805 output, but like I said check the dropout, check the rating of the 3.3V LDO (some have ~6V max voltage ratings, obviously you cannot use the 7V input in this case). Also make sure current ratings for all regulators in your system can handle their respective loads, and if the 7805 is near to it's max load already it's best not to attach the LDO to it.

EDIT: I actually cover some basic things exactly like this, in a 'Seminar' thing I made for my uni's robotics club. Check it out here, in PDF format

My load current for the 3.3 V rail is 546.2 mA. My load current for the 1.8 V rail is 1.154 mA.

Your 1.8V rail is basically just a reference voltage. A voltage divider from a regulated source would work just fine. A 1.8V 1.154 mA load is like a 1500kΩ load, keep that in mind when you make your voltage divider. Efficiency isn't really an issue here. Even at 50% efficiency, that's 4mW.

As for the 3.3V rail, with a 3.6V battery, likely a lithium battery, you want a buck-boost switching regulator, as the voltage range of the battery will be above AND below 3.3V. You want it to regulate down, and then regulate up. Look for one based on your battery chemistry, as there are some that have built in battery protection, to prevent it from draining too low.