You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
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
You can get a simple LCD panel voltmeter from Futurlec, they have various models. The PM128E for $7.90 looks like it would do what you want. I'd suggest using this instead of trying to use the brightness of a bulb to show the output.
For your fan, probably you should just have a separate voltage regulator to supply power to it, since it needs a constant voltage, and not an adjustable one.