Yes, and yes. You can construct the current sense resistor from a fixed resistor of 1.5Ω (to take most of the current) and a 20Ω trimpot (using the wiper and one end) in parallel, for fine adjustment. Alternatively, you can use the 1.5Ω fixed resistor, connect a 100Ω trimpot aross it, and feed the current sense transistor's base from the wiper of the trimpot, like this:
VR1 is a 100Ω trimpot (the value isn't critical). Set it fully counter-clockwise initially and adjust it for the exact current you want. This circuit isn't exact, and variations in supply voltage and LED forward voltage will affect the current somewhat, but not much.
At 500 mA collector current, Q2 will need significant base current. The base current needed by Q2 is equal to the LED current divided by Q2's current gain, but Q2's current gain varies with its collector current! It gets lower as the collector current increases. I've assumed that Q2 has a current gain of at least 20 at 500 mA collector current, so it needs 25 mA base current. We also want a few mA flowing into Q1's collector, so let's say the total current available from RB needs to be 28 mA. Q2's base will sit at around 1.7V, leaving 22.3V across RB, so from Ohm's Law, RB should be about 22.3 / 0.028 which is 796Ω. To guarantee at least that much current, we use the next lower preferred value, which is 680Ω.
For best accuracy, arrange your circuit putting the main current path first. That is the path through the collector and emitter of Q2 and the current sense resistor RS. Then tack the trimpot, the other transistor and the base current supply resistor onto that circuit.
Edit: Other comments following on from Steve's post:
Re the current gain of a BD135/137/139. Fairchild, ON Semiconductor and STMicroelectronics all state a minimum current gain at a collector current of 500 mA, and it is 25. I assumed 20 in my calculations above. But that figure is with Q2 not very strongly saturated. The upshot is that you should use something a lot lower - 330Ω perhaps. But that would dissipate 2W and get pretty hot.
The best solution is to replace Q2 with an N-channel MOSFET, which requires almost no gate current and will saturate very nicely, with a low forward voltage. RB (which should then be called RG) can be increased to around 4k7 to give about 4 mA collector current for Q1. You should add a zener of around 15V (use a 1/2W one) with its cathode to Q2's gate (also Q1's collector) and its anode to 0V, to protect the MOSFET gate from overvoltage if the load becomes disconnected.
You could also use a Darlington transistor for Q2 but this will increase the dropout voltage significantly.
Adding the trimpot to provide easy adjustment of the current increases the dropout voltage slightly because RS is higher than the "perfect" RS. Assuming RS is exactly 1.5Ω it will drop 0.75V at 500 mA. Q2's dropout voltage will be at least 0.25V for a transistor, and somewhat less for a MOSFET, so the total dropout voltage should be around 1V.
Here's the MOSFET version I described.
There are several smaller MOSFETs that would be suitable for Q2 but I only suggested TO-220 devices because they're easier to heatsink.