(i) To radiate the signals with high efficiency.
(ii) Because they are of large wavelength and power radiated by antenna is very small as \(P\alpha \left( l/\lambda \right) ^{ 2 }.\)
(iii) It is done to avoid making over modulated carrier wave. In that situation, the negative half-cycle of the modulating signal is distortion occurs in reception.

Let the receiver be at point A and source be at point B.
Velocity of waves  =3 x 10⁸ m/s
Time to reach a receiver =4.04 ms
= 4.04 x 10^-3s
Let the height of satellite be h_s = 600 km
Radius of the earth = 6400 km
Size of transmitting antenna = h_T
We know that,
 \(\frac { Distance \ travelled \ by \ wave }{ Time } =Velocity \ of \ waves\)
\(\frac { 2x }{ { 4.04\times 10 }^{ -3 } } =3\times { 10 }^{ 8 }\)
\(or \ x=\frac { { 3\times 10 }^{ 8 }4.04\times { 10 }^{ -3 } }{ 2 } \)
\( =6.06\times { 10 }^{ 5 }=606km\)
Using Pythagoras theorem,
\({ d }^{ 2 }={ x }^{ 2 }-{ h }_{ s }^{ 2 }\left( 606 \right) ^{ 2 }-(600)^{ 2 }\)
\( =7236 \ or \ d\ =\ 85.06 \ km\)
So, the distance between source and receiver =2d
=2 x 85.06 = 170 km
The maximum distance covered on ground from the transmitter by emitted electromagnetic waves,
\(d=\sqrt { { 2Rh }_{ T } } \ or \ \frac { { d }^{ 2 } }{ 2R } ={ h }_{ T }\)
\(or \ size \ of \ antenna,\)
\({ h }_{ T }=\frac { 7236 }{ 2\times 6400 } =0.565 \ km=565m\)

Consider the figure given alongside to solve this question.
Suppose,the height of transmitting antenna or receiving antenna in order to cover the entire surface of the earth through communication is h₁ and radius of earth is R.
Then,maximum distance.
 \({ d }^{ 2 }_{ max }=\left( R+{ h }_{ t } \right) ^{ 2 }+\left( R+{ h }_{ t } \right) ^{ 2 }\)
\({ d }_{ max= }\sqrt { 2{ h }_{ t }R } +\sqrt { 2{ h }_{ t }R } =2\sqrt { 2{ h }_{ t }R } \)
\(\because 8{ h }_{ t }R=2\left( R+{ h }_{ t } \right) ^{ 2 }\)
\( \Rightarrow 4{ h }_{ t }R={ R }^{ 2 }+2R{ h }_{ t }+{ h }^{ 2 }_{ t }\Rightarrow { R }^{ 2 }-2{ h }_{ t }R+{ h }^{ 2 }_{ t }=0\)
\(\Rightarrow \left( R-{ h }_{ t } \right) ^{ 2 }=0\Rightarrow R-{ h }_{ t }=0\)
Since space wave frequency is used so, hence only tower height is to be taken into consideration.In three dimensions of earth,6 antenna towers of each of height H_t=R would be used to cover the entire surface of earth with communication programme.

Examples of communication system which use space wave mode are television channel, UHF, VHF, etc.
Maximum distance upto which the signals can be transmitted, d = ?
\(\because \ d=\sqrt { 2hR } \)
\(where, \ R=radius \ of \ the \ earth\)
\(= 6400 \ km\)
\(=6.4\times { 10 }^{ 6 }m\)
\(\therefore \ d=\sqrt { 2\times 80\times 6.4\times { 10 }^{ 6 } } \)
\(=32\times { 10 }^{ 3 }m\)
\(=32 \ km \ \)

The distance travelled by the signal is 5km. Loss suffered in path of transmission = 2dB/km
So, total loss suffered in 5 km = \(-2\times 5=-10dB\)
Total amplifier gain = 10dB + 20dB = 30dB
Overall gain in signal = 30 - 10 = 30dB
According to the question, gain in dB = \(10{ log }_{ 10 }\frac { { P }_{ 0 } }{ { P }_{ 1 } } \)
\(\because \ 20=10{ log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } or\quad { log }_{ 10 }\frac { { P }_{ 0 } }{ P_{ i } } =2\)
Where\({ P }_{ i }=1.01mW\ and\ { P }_{ 0 }\) is the output power.  
\(\because \ \frac { { P }_{ 0 } }{ { P }_{ i } } ={ 10 }^{ 2 }=100\)
\(\\ \Rightarrow { P }_{ 0 }={ p }_{ i }\times 100=1.01\times 100\ or\ { p }_{ 0 }=101mW\)
Thus the output is 101 mW.