(1) There is current flowing(leaking) out the circuit to ground from each of the phases even when there are no faults and the value of this current is several Amps! Mind-blowing!!!

This is correct. Since these lines are very long, when one phase is shorted to ground, every bit of the un-shorted ones leaks a few tiny undetectable mA through air to ground. Just as displacement current inside a capacitor. The sum of all these contributions can reach several Amps going to earth, but at no single point several Amps are flowing, so there is no visible arc flash or similar thing.

(2) If I short a phase to ground. What I will see is the charging current from the other 2 phases coming from ground to my shorted phase through ammeter A2. And what about the current supplied by the utility and that was normally flowing through the phase I have just shorted? What happened to it?

The current supplied by the utility, which in this case would be known as short circuit current, in the case of an ungrounded system (the delta system in the image) with a fault to ground, doesn't have a point to return to the transformer, so it can't flow. That is why in ungrounded systems ammeter A2 is only seeing charging current. If the system was grounded (wye) or resistive grounded (wye with Neutral Grounding Resistor = NGR) ammeter A2 would see the vectorial sum of short circuit current (from the point of fault to the neutral of the wye in the transformer) and charging current (from lines through air to ground to the point of fault)

(3) Will ammeter A1 continue to measure 0 after I short one of the phases? Why?

Yes. Ammeter A1 reads the current (short circuit + charging) through the 3 phases. Since all the current that flows one way, returns through that same way from the current transformer point of view. A1 is unable to detect any missing current and reads 0 Amps

(1) There is current flowing(leaking) out the circuit to ground from each of the phases even when there are no faults and the value of this current is several Amps! Mind-blowing!!!

This is correct. Since these lines are very long, when one phase is shorted to ground, every bit of the un-shorted ones leaks a few tiny undetectable mA through air to ground. Just as displacement current inside a capacitor. The sum of all these contributions can reach several Amps going to earth, but at no single point several Amps are flowing, so there is no visible arc flash or similar thing.

(2) If I short a phase to ground. What I will see is the charging current from the other 2 phases coming from ground to my shorted phase through ammeter A2. And what about the current supplied by the utility and that was normally flowing through the phase I have just shorted? What happened to it?

The current supplied by the utility, which in this case would be known as short circuit current, in the case of an ungrounded system (the delta system in the image) with a fault to ground, doesn't have a point to return to the transformer, so it can't flow. That is why in ungrounded systems ammeter A2 is only seeing charging current. If the system was grounded (wye) or resistive grounded (wye with Neutral Grounding Resistor = NGR) ammeter A2 would see the vectorial sum of short circuit current (from the point of fault to the neutral of the wye in the transformer) and charging current (from lines through air to ground to the point of fault)

(3) Will ammeter A1 continue to measure 0 after I short one of the phases? Why?

Yes. Ammeter A1 reads the current (short circuit + charging) through the 3 phases. Since all the current that flows to the load, returns from that same way from the current transformer (CT) point of view, A1 is unable to detect any missing current and reads 0 Amps. This can be said because the fault is to the right of the CT so charging current flows compensated before it and no short circuit current ever flowed. This explanation is only valid for this specific scenario in the image. If the system was somehow grounded, short circuit current will flow back to the transformer through ground and since an imbalance would be seen in the CT, A1 would not read 0.

(1) There is current flowing(leaking) out the circuit to ground from each of the phases even when there are no faults and the value of this current is several Amps! Mind-blowing!!!

This is correct. Since these lines are very long, when one phase is shorted to ground, every bit of the un-shorted ones leaks a few tiny undetectable mA through air to ground. Just as displacement current inside a capacitor. The sum of all these contributions can reach several Amps going to earth, but at no single point several Amps are flowing, so there is no visible arc flash or similar thing.

(2) If I short a phase to ground. What I will see is the charging current from the other 2 phases coming from ground to my shorted phase through ammeter A2. And what about the current supplied by the utility and that was normally flowing through the phase I have just shorted? What happened to it?

The current supplied by the utility, which in this case would be known as short circuit current, in the case of an ungrounded system (the delta system in the image) with a fault to ground, doesn't have a point to return to the transformer, so it can't flow. That is why in ungrounded systems ammeter A2 is only seeing charging current. If the system was grounded (wye) or resistive grounded (wye with Neutral Grounding Resistor = NGR) ammeter A2 would see the vectorial sum of short circuit current (from the point of fault to the neutral of the wye in the transformer) and charging current (from lines through air to ground to the point of fault)

(3) Will ammeter A1 continue to measure 0 after I short one of the phases? Why?

Yes. Ammeter A1 reads the current (short circuit + charging) through the 3 phases.

(1) There is current flowing(leaking) out the circuit to ground from each of the phases even when there are no faults and the value of this current is several Amps! Mind-blowing!!!

This is correct. Since these lines are very long, when one phase is shorted to ground, every bit of the un-shorted ones leaks a few tiny undetectable mA through air to ground. Just as displacement current inside a capacitor. The sum of all these contributions can reach several Amps going to earth, but at no single point several Amps are flowing, so there is no visible arc flash or similar thing.

(2) If I short a phase to ground. What I will see is the charging current from the other 2 phases coming from ground to my shorted phase through ammeter A2. And what about the current supplied by the utility and that was normally flowing through the phase I have just shorted? What happened to it?

The current supplied by the utility, which in this case would be known as short circuit current, in the case of an ungrounded system (the delta system in the image) with a fault to ground, doesn't have a point to return to the transformer, so it can't flow. That is why in ungrounded systems ammeter A2 is only seeing charging current. If the system was grounded (wye) or resistive grounded (wye with Neutral Grounding Resistor = NGR) ammeter A2 would see the vectorial sum of short circuit current (from the point of fault to the neutral of the wye in the transformer) and charging current (from lines through air to ground to the point of fault)

(3) Will ammeter A1 continue to measure 0 after I short one of the phases? Why?

Yes. Ammeter A1 reads the current (short circuit + charging) through the 3 phases. Since all the current that flows one way, returns through that same way from the current transformer point of view. A1 is unable to detect any missing current and reads 0 Amps