The diode is in place to protect the transistor from reverse \$V_{BE}\$ breakdown. If you reverse-bias the base-emitter junction by taking the transistor's emitter to ~\$6\rm{V}\$ more positive than the base, it will breakdown and begin conducting.
This reverse breakdown damages the base-emitter junction, causing a degradation in \$h_{FE}\$.
By placing a diode as shown, the reverse-bias voltage is limited to ~\$0.7\rm{V}\$; attempts to apply more voltage will be futile, since the diode will conduct a lot of current and prevent increased voltage. This protects the transistor's base-emitter junction.
I am having trouble understanding the three operation modes of a transistor.
When we talk about the modes of operation of a transistor, we're usually talking about cut-off, forward-active, and saturated operation.
The rest of your question seems to be about the different fundamental amplifier configurations, rather than the operating modes, so that's what I'll answer about.
Consider the CE mode. The collector and Emitter both are negative (for npn)
Check your diagrams again. For an NPN CE stage, the base and collector are both biased at higher potential than the emitter.
then how can we reverse bias them?
For a CE stage, the base-emitter junction should be forward biased; the base-collector junction is reverse biased. This is achieved by biasing the collector at a higher potential than the base. This is exactly what's shown in the diagram you posted.
Why is the common pin (the base , the emitter , and the collector respectively) grounded? what is the significance of grounding them?
They aren't necessarily grounded. They are connected to some potential that is equivalent to ground in the ac equivalent circuit. Particularly for common-base stages or PNP common-emitter stages, this is usually not the same as the circuit ground.
The significance is that this node is "common". A node that is used in common between the other nodes to define their potentials. The fact that the emitter is connected to the common node is why we call a common-emitter stage a common emitter stage.
How does a CE mode amplify more than a CB mode ?
A common emitter stage has voltage and current gain. A common base stage is essentially a unity gain current buffer. You need to study the common emitter stage to understand why it has voltage and current gain, and study the common base stage to understand why it is a unity gain buffer. Once you understand those two things, you'll understand why the one stage has more gain than the other.
Best Answer
Voltage is the electric potential difference between two points. A single point doesn't have an absolute voltage, only relative voltages compared to other points in the circuit. Of course, often there is an implicit reference point, the "ground" (either something actually grounded, or just the negative or middle point of the power supply marked as "the ground").
So, the voltage "at the base" is either meaningless, or implicitly means the voltage between the base and the ground, and similarly for the voltage at the emitter.
But note that in the quote there is a comparison between two points: the voltage between the points "at the base" and "at the emitter" can be measured directly, without needing a third reference point. This is the voltage that matters, regardless of what the voltages at the base and emitter are compared to other points in the circuit. E.g. in an emitter follower, neither might be grounded, so the voltages against the ground are not relevant.
What the voltage between the base and the emitter actually is, depends totally on the surrounding circuit. If both are tied to ground (or whichever one point), the voltage between them is zero, and the transistor is "off". If the base is driven at least 0.7 V (or so) higher than the emitter, the transistor is "on", and the resulting current will be such that the base-emitter voltage stays around 0.7 V (but really it depends on the current). Unless you connect a voltage source directly across the transistor, of course, in which case the voltage and current would be higher, for a short moment.
It may help to play with it. See the linked circuit in the Falstad simulator. You can vary the voltage that goes to the base (through the resistor) with the slider. (The arrows are current meters, not diodes.)
As for the voltage "remaining after all voltage drops", well, in a way. In the above circuit, the voltage between the transistor base and emitter is whatever the voltage source to the base gives, minus what's lost on the base resistor. That's probably not very helpful in practice, though.