Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way?
I was directed to the following YouTube video and it contains the best explanation of the phenomenon that I am seeing.

What is ammeter burden voltage, and why you should care (YouTube)

My extra tests seemed to back up the idea in the video (that the meter's internal resistance is around 9-10Ohms).

Specifically when I measured the voltage during load across the 10 Ohm resistor and got 2.2V.

2.2 / 10 Ohm = 220mA


However, I know when I measure that I only see 157mA of current. So...

220mA - 157mA = 63mA (which are missing somewhere).


I plug in the missing 63 in the following way:

.8V (voltage drop) / .063 = 12.69 Ohms


The point here being that I still have ~10-12Ohms of resistance I can't account for (which are assumed to be in the meter itself).

Last Point of Data Which Seems to Confirm Internal Resistance Theory

Another poster pointed out a special port on my meter (marked 20A) which has a smaller burden (around 1 Ohm).  I plugged my positive probe into that port and I measured 270mA on the circuit -- very close to the 300mA I was expecting.



In answer to my other questions, the video also shows the meter data sheet items that will show the internal resistances when purchasing an multimeter.

EDIT

I kept raising the resistors (but continue to see the extra ~10 Ohms of resistance).
For example, I just put a 330 Ohm resistor and measured 8.81mA of current.
If I calculate the value using Ohm's law I get:

3V / 330 Ohms = .0090A (9.0mA)


If I adjust my calculation considering that my meter is causing an additional 10 Ohms of resistance I get almost the exact value that I saw when I measured.

3V / 340 Ohms = .00882 (8.82mA)


Update

There was also some question about whether or not the AA battery could even provide 300mA of current due to internal resistance and I wanted to close the loop on that.  

I looked up the duracell AA datasheet and although it does not specifically state max current the battery can provide it does show the following graph:



You can see that they do show it providing up to 1000mA (1 amp) of current for a little over an hour so I believe from this we can infer that the battery can actually supply current at these higher levels (300mA and more).