Electronic – Do I need a gate resistor and/or a pull-down resistor for a TC4420 (or any gate driver) driving a MOSFET

The gate resistor is normally used with power MOSFETs to limit the magnitude of the current spike that can occur when a driver switches fast and has to charge or discharge a large gate capacitance. With a super fast driver with high drive capability the current spikes to directly drive a large gate capacitance can be measured in the Amps. A small resistor, such as the 100 ohm resistor that you show, can nicely reduce these current spikes to a reasonable level. An a reasonable controlled level current surge is not going to cause coupling and upset to nearby circuits.

Another reason to use the series resistor in the gate circuit is to keep the current surge within the max drive spec for a whimpy driver circuit such as a micro controller GPIO pin.

The MOSFET that you selected has quite low input capacitance (~75 pF) such that the MCU pin can easily drive the gate without the resistor to limit the current.

You mention using this part because it has a "Vgs" of one volt. Do not fool yourself with this spec. A Vgs at 1V only allows a typical drain current of 250uA to flow. You didn't specify the current requirement for the GPS receiver so limited comment can be offered. However you should study the data sheet carefully. Using the graph copied and shown below you will note that with the Vgs of 3.3V as offered from your microcontroller drive pin places the operating mode somewhere between the lower two curves on the chart. You can reasonably expect the available drain current to be be less than 300mA and to be largely limited by the drain-source resistance of the part. Unless your GPS unit runs on very low current (few mA) it is likely that the voltage level seen at the GPS unit supply pin will be less than the 3.3V that you would like to see.

The gate resistor on a MOSFET is really there to protect whatever is sourcing the current. Much like a discharged capacitor, the gate will initially look like a short to ground when voltage is first applied. A MOSFET with a very large gate capacitance can sink a very large amount of current for a short period of time. If you're driving the gate with, say, a MCU pin, it's usually a good idea to put a small resistor to reduce that current surge to a value the MCU can handle. If your MCU pin can handle, say, 20mA and you're driving 3.3V into the gate, then you choose a resistor that limits the current to 20mA at 3.3V:
$$R_{gate}=\frac{V}{I}=\frac{3.3V}{20mA}=165Ω$$

The digital output pins of most MCUs are current limited already, so this isn't strictly necessary. But why pound on the output drive circuitry to the point that the current limiter kicks in?

Incidentally, MOSFET drivers are ICs specifically made to drive a large amount of current into the gate of a MOSFET so as to turn on the MOSFET as quickly as possible. The MOSFET turns on quicker and the switching losses are less.

All of that said, I believe you have another issue with your circuit. I suspect 3.3V is not enough to turn on two UV LEDs in series. The voltage will need to be at least as high as 2x the foward voltage of one of those LEDs. As it is, I don't think the LEDs will turn on.