D-S junction should show ohmic resistance on a JFET, with gate open.
Gate to either terminal shows open circuit, or very high resistance.
If you don't need reliable identification of the D and S terminals (which is which?) or operating parameters an ohmmeter/multimeter is sufficient.
No.
Shorting Gate to Source is equivalent to 0V for Vgs. This is the way one tests for Idss.
If the bf245 is symmetrical, then shorting the Gate to Drain should also give the same test result for Idss @ 0Vgs.
If Idss is very high, then testing for Idss @ near 10Vds will rapdily overheat the jFET.
Start with a much lower voltage. Or insert a 1k0 resistor in the Source lead and short the gate to the other end of that 1k0 resistor. This is an Nchannel jFET and needs +ve voltage to the Drain. -ve voltage to the other leads.
Leaving the Gate open leaves whatever charge is on the gate. This is likely to be not zero volts Vgs. This could show as an open circuit between Drain and Source.
If you short Drain to Source, that is a different arrangement. I think a jFET in this arrangement shows current flow in both directions, i.e. a low ohms test result. Can someone confirm this for D shorted to S?