$$
c=3times 10^8frac{textrm{m}}{textrm{s}}
$$

$$
c=lambda f
$$

from where we can express the the wavelength as

$$
lambda=frac{c}{f}=frac{3times 10^8}{5.7times 10^{14}}
$$

Finally, we have that

$$
boxed{lambda=5.26times 10^{-7}textrm{ m}}
$$

$$
c=lambda f
$$

from where we can express the the wavelength as

$$
lambda=frac{c}{f}=frac{3times 10^8}{8.2times 10^{14}}
$$

Finally, we have that

$$
boxed{lambda=3.7times 10^{-7}textrm{ m}}
$$

$$
c=lambda f
$$

from where we can express the frequency as

$$
f=frac{c}{lambda}=frac{3times 10^8}{2.2times 10^{-2}}
$$

Finally, we have that

$$
boxed{f=1.4times 10^{10}textrm{ Hz}}
$$

$$
v=frac{c}{sqrt{K}}=frac{299,792,458}{sqrt{1.00054}}
$$

Which gives that in air the speed of electromagnetic waves is

$$
boxed{v=299711547, frac{textrm{m}}{textrm{s}}}
$$

$$
v=frac{c}{sqrt{K}}=frac{3times 10^8}{sqrt{1.77}}
$$

Which gives that in water the speed of electromagnetic waves is

$$
boxed{v=2.25times 10^8, frac{textrm{m}}{textrm{s}}}
$$

$$
v=frac{c}{sqrt{K}}
$$

From here one can express the dielectric constant as

$$
K=frac{c^2}{v^2}
$$

which after we plug in the values becomes

$$
K=frac{3^2times 10^{16}}{2.43^2 times 10^{16}}
$$

$$
boxed{K=1.52}
$$

$$
c=lambda f
$$

from where we can express the frequency as

$$
f=frac{c}{lambda}=frac{3times 10^8}{1.5times 10^{-2}}
$$

Finally, we have that

$$
boxed{f=2times 10^{13}textrm{ Hz}}
$$

$$
fpropto frac{1}{lambda}
$$

This means that the higher frequency waves will have shorter wavelengths and vice versa. Therefore the channels from 2 to 6 will require longer antennas.

$$
v=frac{c}{sqrt{K}}
$$

From here one can express the dielectric constant as

$$
K=frac{c^2}{v^2}
$$

which after we plug in the values becomes

$$
K=frac{3^2times 10^{16}}{1.98^2 times 10^{16}}
$$

$$
boxed{K=2.3}
$$