$textit{b.}$, Sound of clarinet is produced by reeds.

$textit{c.}$, Sound of tuba is produced by the lips of the player.

$textit{d.}$, Sound of violin is produced by string.

$textit{b.}$ reeds

$textit{c.}$ lips of the player

$textit{d.}$, string

$f_{1} = 370, mathrm{Hz}$

$f_{2}$ , $f_{3}$, $f_{4}$ $= ?$

Harmonics are given by the product of a whole number and fundamental ($f_{1}$):

$$
f_{2} = 2 f_{1}
$$

$$
f_{3} = 3 f_{1}
$$

$$
f_{4} = 4 f_{1}
$$

When we put known values into the previous equations:

$$
f_{2} = 2 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{2} = 740, mathrm{Hz}}
$$

$$
f_{3} = 3 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{3} = 1100, mathrm{Hz}}
$$

$$
f_{4} = 4 cdot 370, mathrm{Hz}
$$

$$
boxed{f_{4} = 1480, mathrm{Hz}}
$$

$$
f_{3} = 1110, mathrm{Hz}
$$

$$
f_{4} = 1480, mathrm{Hz}
$$

$L = 2.40, mathrm{m}$

$textit{a.}$, $f = ?$

Frequency is given by:

$$
f = dfrac{v}{lambda}
$$

Therefore, we need to determine wavelength since the speed of sound is known $v = 343, mathrm{m/s}$:

$$
begin{align*}
lambda &= 4L\
&= 4 cdot 4.20, mathrm{m}\
&= 9.60, mathrm{m}\
end{align*}
$$

Frequency will, finally, be:

$$
f = dfrac{343, mathrm{m/s}}{9.60, mathrm{m}}
$$

$$
boxed{f = 35/7, mathrm{Hz}}
$$

$f _{2} = 1.40, mathrm{Hz}$

First we need to determine frequency, and since both pipes are played at the same time, it will now be:

$$
f ‘= f – f_{2}
$$

When we put known values into the previous equation we get:

$$
f = 35.7, mathrm{Hz} – 1.40, mathrm{Hz} = 34.3, mathrm{Hz}
$$

Since length is given by:

$$
L’ = dfrac{lambda}{4}
$$

We have to determine wavelength first:

$$
lambda = dfrac{v}{f}
$$

When we put known values into the previous equation we get:

$$
lambda = dfrac{343, mathrm{m/s}}{34.3, mathrm{Hz}} = 10, mathrm{m}
$$

Finally:

$$
L = dfrac{10}{4}
$$

$$
boxed{L’ = 2.50, mathrm{m}}
$$

Therefore difference will be:

$$
L – L’ = 2.50, mathrm{m} – 2.40, mathrm{m} = boxed{0.10, mathrm{m}}
$$

$textit{b.}$, $0.10, mathrm{m}$

Therefore the frequency of the other fork is either :
$$
392-3 = 389hspace{10mm}text{or}hspace{10mm}392+3 = 395
$$