About 5V, 500mA coming out of R_load now. At this time, when I gradually reduce the R_load, the voltage and current increase. In this situation, I need to limit R_load to 5V,500ma, what should I do?

 

First of all I would put the current source between the battery and the voltage regulator -just guessing that a fixed voltage is desired when current is not being regulated. The I would change R5 to 10 ohms. That will limit your current to 500 ma. The resistor will have to be able to dissipate nearly 3 watts (2.5W).

 

For a 500mA limit, R1 = 5V / 500mA = 10Ω.
For a 5V output, the LM7805 needs no resistors, just connect the GND terminal to circuit ground/common.

You need to reverse the positions of the two circuits so that the current-limit circuit feeds the constant-voltage circuit.

Note that the current limit circuit drops 5V across the sense resistor before it limits.
For a smaller drop (1.25V), use an LM317 for the current limit (circuit below):
With that, the maximum R3 dissipation is 625mW.
The yellow trace is the varying load resistor value with time.
The output voltage (red trace) stays at 5V until the load current (green trace) reaches 500mA, at which point the voltage starts to drop to maintain a constant 500mA through the load.

Edit: You can also, of course, use the LM317 for both the current limit and the 5V output.

 

About 5V, 500mA coming out of R_load now. At this time, when I gradually reduce the R_load, the voltage and current increase. In this situation, limit R_load to 5V,500ma. So with R_5 taking 2.5w and using only 0.25w of resistance, is there a better way than just connecting 10*(1 ohm) resistors?

 

is there a better way

As I previously stated, and showed in my circuit, you need to put the current-limit circuit before (to the left of) the voltage regulating circuit.
Otherwise you are seeing the drop of the current-limit circuit at the output even before it limits.
Did you not understand that?

 

is there a better way than just connecting 10*(1 ohm) resistors?

What exactly are you trying to achive? To answer your question: You could use five 2 Ohm resistors in series or five 2 Ohm resistors in series, one 10 Ohm resistor or two 20 ohm resistors in parallel. It all depends on what you have available.

 

About 5V, 500mA coming out of R_load now. At this time, when I gradually reduce the R_load, the voltage and current increase. In this situation, limit R_load to 5V,500ma. So with R_5 taking 2.5w and using only 0.25w of resistance, is there a better way than just connecting 10*(1 ohm) resistors?

Yes there's a far better way than your circuit, as already stated in post#3, but what current do you want to limit to??

 

Is there a better way to divide 2.5w by connecting resistors

 

Is there a better way to divide 2.5w by connecting resistors

Putting equal value resistors in series or parallel to get the desired resistance will divide up by power by the number of resistors.
For example, putting five, 2Ω resistors in series to get 10Ω total, will generate 0.5W max dissipation in each resistor,