what is exactly 'gain and loss' is and how is it used?
"Gain" is just the ratio of output strength to input strength.  For example, if a amplifier increases a voltage by a factor of 4, then the amplifier can be said to have a gain of 4.

"Loss" is a bit more ambiguous term.  It can mean just the reciprocal of gain.  It is usually applied when a signal is attenuated instead of amplified.

"Loss" often carries the connotation that the reduction in signal strength is unintentional.  For example, the signal out of a long cable will be weaker than what went in.  The ratio of input to output is the loss in the cable.  Note that this is the reciprocal of gain.

To add more confusion, note that both gain and loss are often expressed logarithmically. That may sound unnecessarily complicated, but it actually simplifies working with gains and losses in many contexts.

Decibels (dB) are commonly used to express gain and loss logarithmically.  A Bel is simply the log10 ratio of output to input power.  However, that would mean a change of 1 corresponds to a power change of 10.  That was deemed too large a step for common use, so the deciBel, abbreviated dB, is used instead.  dB is defined as

&nbsp; &nbsp; dB = 10 log10(OutPower/InPower)

One advantage of expressing power gain (and loss) logarithmically is that sequential gains just add.  What would be a multiply with linear gains is now addition with logarithmic gains.

For example, let's say you want to calculate the overall gain from a radio transmitter to a receiver.  The cable from the transmitter to the antenna has a loss of 5 dB, the antenna has a gain of 3 dB in the direction of the receiver, and the signal is attenuated another 39 dB due to the distance from the antenna to the receiver.  Overall you have a system with a gain of (-5 dB)+(3 dB)+(-39 dB) = -41 dB.  Or, you could say it has a loss of 41 dB.

Note how dB is defined in terms of power.  Sometimes you are dealing with power directly, like in the RF example above.  Other times we are more interested in the voltage gain.  Note that power is proportional to the square of voltage, all else held constant.  dB for voltages is therefore:

&nbsp; &nbsp; dB = 10 Log10(Vout2/Vin2)

From that follows:

&nbsp; &nbsp; dB = 10 Log10((Vout/Vin)2)

&nbsp; &nbsp; dB = 20 Log10(Vout/Vin)

Since you're in 12th grade (high school senior), you should have been exposed to logarithms enough to understand the above.

Of course the power of a signal is only proportional to the square of the voltage.  The actual power also depends on the impedance of that signal.  However, when we only really care about the voltage, we just ignore that part to be able to use the handy dB units.

For example, let's say a audio amplifier takes "line level" audio signals in, and can (adjustable via the volume knob) amplify the voltage by up to 10 before that is fed into the power amp that actually drives the speakers.  In addition, this preamp also has a microphone input that adds a fixed voltage gain of 1000 before the line input.

In this case, the microphone preamp can be said to have a gain of

&nbsp; &nbsp; 20 log10(1000) = 60 dB

and the basic preamp can have a gain up to 

&nbsp; &nbsp; 20 log10(10) = 20 dB

Overall, from microphone input to output, this preamp can be said to have a gain of (60 dB)+(20 dB) = 80 dB.

We can work this backwards to get the resulting voltage gain.  1080/20 = 104 = 10,000.  That checks, since the voltage gains multiplied out (1000 x 10) come out to the same value.