For current-mode drivers, why is it easier to control output impedance ?
With the voltage-mode drivers, the resistance of the output transistors is in series with the source termination. With the current-mode drivers, the resistance of the output transistors is in series with a constant current source making it irrelevant except in lowering the output compliance voltage.
Note that in both cases, it is not absolutely necessary to use source termination which would halve the current required for the current-mode driver and halve the voltage required by the voltage-mode driver but the series resistance of the transistors in the voltage-mode driver would still be a problem.
For voltage-mode drivers, why "0.25 to 0.5" of the current needed for a given output voltage swing?
That is incorrect. For a given output voltage, the voltage-mode drivers only require 1/2 the current because they are driving twice the output resistance from the double termination but they also only produce 1/2 the voltage.
What is the purpose of the two opamps in the voltage-mode driver shown the first post?
The operational amplifiers just represent the regulated supply voltages. Note that the second operation amplifier is providing twice the output voltage because half is lost in the source termination which is inexplicably not shown.
How do the two opamps contribute to current amount reduction?
They do not. That is a function of the double termination (not shown) used with the voltage output drivers. They author may have been suggesting that a second supply voltage of twice the value is necessary to get the same output as the current-mode driver.