1A Voltage Regulator – Maximum or output

What you need is an LDO (Low DropOut) regulator. The Micrel MIC2937 has a dropout of only 370 mV at 750 mA out.

You'll have to check the output voltage of the adapter at maximum current. Connect a 16 \$\Omega\$/10 W resistor and measure the voltage. You can combine resistors you have in your box, keeping in mind that P = V\$^2\$/R. The voltage should be higher than 12.4 V.

The MIC29372 (same datasheet) is a version with adjustable output, which may be of help if the output voltage isn't high enough for 12 V out.

edit
If the voltage is too low, and you really want to use the adapter, you can replace the rectifier diodes with Schottky types. That will give you an extra volt output.

edit 2 (regarding the LM317 hli also proposes)
If it's a 12 V adapter it will deliver about 12 V under load. Saying that the unloaded voltage is high enough for the LM317 is meaningless; you do want to use it, don't you? 12 V, even a bit higher is insufficient for the LM317. Don't design for unreliability and use an LDO.

edit 3 (2012-06-14, re your comment) If the adapter is only going to be used only for powering a relay you don't need a regulator. The relay will only show a small load to the 12 V, so it may well be a few volts higher (unloaded transformers output a higher than nominal voltage). This relay is rated at 12 V, but has a maximum voltage 1.7 \$\times\$ that, so that's 20.4 V, which should be safe. Nominal power is 360 mW, and that will be somewhat higher at 15 V for instance (560 mW). If the voltage at 30 mA load is much higher than the 12 V you could still use a regulator. This will also consume some power, but the sum of regulator and relay power would be less than the relay alone.

Also nice about it is Must operate voltage, which is maximum 75 % of rated voltage. That means it will still work if the adapter's voltage would drop a few volts. The Micrel regulator will drop less than 200 mV at 30 mA, if we go for the adjustable, and choose 10 V out, then we will have proper regulation down to 10.2 V in, and the relay will only consume 250 mW instead of 360 mW. So we win twice.

1. R30 limits the charging peak current to the capacitor somewhat, but at 1 ohm it still allows for 12 A, so of little use there. Also would limit the current through the zener if there are peaks above 20 V.

2. The larger capacitors work less well at higher frequencies, and that's where the smaller ones take over.

3. 470 µF in a non-polarized version would be expensive, but there would be nothing against it. All large capacitors are polarized.

4. On the output it would give an extra load for the 7805 to charge it. On the input too for the battery, but that can deliver more than enough current.

5. No, except that they're larger.

6. Keep in mind that the input-output difference for the 7805 is about 5 V (12 V - 1 V for the diode and 1 V for R30 - 5 V out) and that at 1 A out the regulator will dissipate 5 W, so it will need considerable cooling (sizable heatsink) if you want to draw that much current.