They don't say, so you don't know. If the chain of cheap copies stretches back to something industry standard, it's 10M-ohm or so. If not -- you don't know.
You may be able to figure this out with a largish capacitor -- if you connect a fairly accurate 10uF ceramic cap to the leads, charge it up with a battery, then disconnect the battery, the voltage should decay exponentially. Plot voltage vs. time on semilog graph paper, and the slope will tell you the time constant. Then use tau = 1/(R*C) to calculate R.
Alternately, get your hands on a string of 10 1M-ohm resistors and measure a battery's voltage through the string -- if the string cuts the measured voltage in half, it's a 10M-ohm meter.
Manual.pdf
No, it is not. Measure the signal generator voltage with no load, then with a resistor across the outputs. A resistor that cuts the voltage by a factor of two should (if the meter is decently built) equal the generator output impedance.
It's a good idea to check the output impedance over frequency, as it may vary. It shouldn't, but it's an instrument without specifications.
Unless you're living in abject poverty the best way to find out the input impedance of your meter and the output impedance of your signal source is to throw the cheap stuff away and buy something that actually specifies these things and is made by someone who actually builds to specification.
I understand that this may be beyond your means, but unless you're doing all the messing around for fun and for free, or you're just barely getting by financially and calibrating your own instruments is part of pulling yourself up by the bootstraps, then you'll probably come out ahead in the long run by spending more money up front and getting decent test equipment.
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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
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