If power requirements are low (tens of millamps) I tend to use linear regulators. It doesn't matter whether you go from 12V to 3.3V immediately, or via 5V, the efficiency is the same (and rather bad). Make sure, however, if you use a regulator to go from 5V to 3.3V to use an LDO; the voltage difference is too small for standard regulators.
If you want to use SMPSs you'll have to look at the efficiency. This is in general lower for lower output voltages, and also if the Vin/Vout ratio gets higher. Let's say you go from 12V to 5V at 90%, from 12V to 3.3V at 80% and from 5V to 3.3V at 95%. So going from 12V to 3.3V in 1 step will be 80% efficient, if you do this in 2 steps (via 5V) your overall efficiency will be 90% * 95% = 85.5%. You'll have to make the calculation for your specific regulator.
I agree with others that switchers are a better choice in terms of efficiency, but they can be somewhat complicated to deal with if you're inexperienced, and there can be lots of weird effects that aren't immediately obvious (precharge sinking, beat frequencies, etc.) that can make life difficult. Assuming you've figured out your power dissipation and know how much current each rail can deliver, if the linears will work for you, stick with them (at least for the first pass).
If you're trying to achieve a variable-amplitude square wave output on your adjustable rail, the chopping may introduce noise into the main 24V rail, which could show up on the other rails. You may want to have an LC filter between the main 24V rail and the regulator input to provide high-frequency isolation, and will probably need extra capacitance on the adjustable regulator output (bulk electrolytic as well as low-impedance ceramic) if you expect the square wave edges to be sharp.
1, 5) There are some dangers with your scheme.
Power dissipation in the linear regulators will be
\$(V_{out} - V_{in}) \cdot I_{out} \$
which is significant, especially for the lower output rails. 78xx-type regulators have built-in thermal protection around 125°C, and (without heatsinking) a junction-to-air thermal resistance of 65°C/W. Your thermal management will be challenging.
Another potential problem - if the series-pass element in any of your low-voltage regulators fails or gets bypassed (shorted), you'll present the full 24V input to the output. This could be catastrophic to low-voltage logic. You should protect your low-voltage rails with SCR crowbars that can sink enough current to put the DC/DC brick into current limit and collapse the 24V rail (they'll need big heatsinks too). Fuses are unlikely to be good protection since the 24V brick likely isn't stiff enough to generate the \$I^2 \cdot t\$ needed to blow a fuse.
2) Whatever floats your boat.
4) Meters aren't huge loads. Just use one of your rails.
3) Correct - all regulators have headroom requirements. If you want the maximum 24V out, you'll need a direct connection, and will have to rely on whatever intrinsic protections the brick will provide you.
Best Answer
Simple answer: You would get 5V in your case, but referenced to -12V instead of ground. In other words, you would have -7V, not -5V. In addition, the regulator would only source current onto the -7V rail, not sink it as would be expected for a negative voltage.
If you want to run some circuitry from 5V between -7V and -12V (the -12V will be the ground for this circuitry), then you can use a positive regulator as you described. If however you want to run some circuitry between ground or higher and -5V, then you need the negative regulator.