(1) Thanks to this lecture note by MIT, it seems like is to indicate that the path from the point to the point of the amplifier has current. And the point of ? Well, I’m not sure: a slide from University of Kansas.
(3) Virtual ground. Reference: Khan Academy.
(4) Thanks to slide 8 of Georgia Tech, this is Binary Weighted Input DAC:
(5) It is given from (3) that ideal operational amplifiers have . So:
(8) The circuit in Fig 3 is more numerically efficient than the circuit in Fig 2 as the number of bits becomes big, Fig 2 will require resistors with large impedance, which can be costly and error-prone. Yet, the circuit in Fig 2 requires less number of resistors and it is easier to be implemented, which can be ideal of a small number of bits.
The slide 11 and 16 of Georgia Tech also suggests the speed of conversion in Fig 2 will be faster than in Fig 3. I guess this is due to the fact that the current in Fig 3 must go through a resistor R before it can get to the resistor 2R (except for the MSB).