$textbf{color{#c34632}(a)}$
$$
text{Speed of Sound} = dfrac{text{Distance travelled by sound}}{text{Time taken}}
$$

$$
text{Speed of Sound} = dfrac{515text{ m}}{1.5text{ s}}approx 343.3text{ m/s}
$$

$textbf{color{#c34632}(b)}$
$$
T = dfrac{1}{f} = dfrac{1}{436text{ Hz}}approx 2.3times10^{-3}text{ s}
$$

$$
textbf{color{#c34632}(c)}
$$

$$
lambda =dfrac{v}{f} = dfrac{343.3text{ m/s}}{436text{ Hz}}approx 0.79text{ m}
$$

$$
text{color{#4257b2}$$textbf{color{#c34632}(a) }343.3text{ m/s}$$

$$textbf{color{#c34632}(b) }2.3times10^{-3}text{ s}$$

$$textbf{color{#c34632}(c) }0.79text{ m}$$}
$$

$text{underline{Given:}}$

$d=465 text{m}$      $t_{echo}=2.75 text{s}$

$lambda=0.750 text{m}$     $upsilon=343 dfrac{text{m}}{text{s}}$

$textbf{a)}$ We need to determine the speed of sound of the hiker’s voice in air.

To determine that, we use the speed of wave (sound) equation, which is given by:

$$
begin{align*}
upsilon &=dfrac{d}{t} tag{$dfrac{text{Distance}}{text{Time taken}}$} \
end{align*}
$$

As we know, an echo travel bounces off a thing, and comes back the same distance, hence above equation must be rearranged,

$$
begin{align*}
upsilon &=dfrac{2d}{t} tag{Equation 1.}\
end{align*}
$$

Plug in values into equation 1 and solve for speed of sound,

$$
begin{align*}
upsilon &=dfrac{2 cdot (465 text{m})}{2.75 text{s}} tag{Plug in values.}\
upsilon &=dfrac{ 930 text{m}}{2.75 text{s}} \
upsilon &=338.18 dfrac{text{m}}{text{s}} \
end{align*}
$$

$textbf{b)}$ Frequency is defined as:

$$
text{Frequency}=dfrac{text{Speed of sound}}{text{Wavelength}}=dfrac{upsilon}{lambda}
$$

Plug in values into above relation and solve for frequency:

$$
begin{align*}
f =dfrac{ 338.18 dfrac{text{m}}{text{s}} }{0.750 text{m} }=450.9 text{s}^{-1} quad text{or} quad 450.9 text{Hz} \
end{align*}
$$

$textbf{c)}$ Mathematically, the period is the reciprocal of the frequency.

$$
T=dfrac{1}{f}.
$$

Substitute values into period relation and solve,

$$
begin{align*}
T &=dfrac{1}{450.9 text{Hz}} tag{Plug in values.}\
T &=dfrac{1}{450.9 text{s}^{-1}} tag{$1 text{Hz}=1 text{s}^{-1}$.}\
T &=0.00221 text{s} \
T &=2.21 cdot 10^{-3} text{s} tag{In scifient notation.}\
end{align*}
$$

$textbf{a)} upsilon =338.18 dfrac{text{m}}{text{s}}$

$textbf{b)} f =450.9 text{Hz}$

$textbf{c)} T=2.21 cdot 10^{-3} text{s}$

In case when we have one sinusoidal wave and this wave moving at a fixed wave speed, the wavelength of this wave will be inversely proportional to the frequency of the this wave. In math terms:
$$
lambda=dfrac{upsilon}{f}.
$$

Note that the waves with lower frequencies have longer wavelengths and for higher frequencies vise versa apply.

The answer is: you should shake the rope at a lower frequency for increase the wavelength of waves in a rope.

$text{underline{Given:}}$

$f=3.5 text{Hz}$

$lambda=0.7 text{m}$

Our task is to find the value of the speed of a periodic wave disturbance from at the given values of the frequency and wavelength.

To find that, the best way is using the speed of wave equation:

$$
text{Speed of wave}=text{Frequency} cdot text{Wavelength}
$$

In math terms,

$$
upsilon=f cdot lambda.
$$

Plug in values into above relation and solve for the speed of a periodic wave,

$$
begin{align*}
upsilon &=(3.5 text{Hz}) cdot (0.7 text{m}) tag{Plug in values.}\
upsilon &=(3.5 text{s}^{-1}) cdot (0.7 text{m}) tag{Convert Hz to seconds.}\
upsilon &=2.45 dfrac{text{m}}{text{s}}\
end{align*}
$$

$upsilon =2.45 dfrac{text{m}}{text{s}}$

$$
text{color{#4257b2}We know that $$v = lambdacdot f$$

$ $

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

$$
lambda = dfrac{15text{ m/s}}{text{6 Hz}}=2.5text{ m}
$$

$$
text{color{#4257b2}
$$lambda = 2.5text{ m}$$}
$$

$text{underline{Given:}}$

$lambda=1.20 text{cm}$

We have to find value of the speed of propagation of the wave. Initial step is to convert values of the wavelength from the cm to m.

Since one centimeter is one hundredth of a meter, we divide the value of the wavelength expressed in centimeters by 100 to get the value expressed in meters.

$$
lambda=dfrac{1.20 text{cm}}{100}=boxed{0.012 text{m}}
$$

Next step is to find the value of time needed to generate one pulse. We have 5 pulses are generated every 0.100 second, time needed for generating one pulse is:

$$
t=dfrac{0.100 text{s}}{5}=boxed{0.02 text{s}}
$$

Using the speed of wave equation, we get required value.

$$
begin{align*}
upsilon &=dfrac{lambda}{t} tag{Speed of wave equation.}\
end{align*}
$$

Plug in values into speed of wave equation and solve,

$$
begin{align*}
upsilon &=dfrac{0.012 text{m} }{0.02 text{s}} tag{Plug in values.}\
upsilon &=0.6 dfrac{text{m} }{text{s}} \\
end{align*}
$$

$upsilon =0.6 dfrac{text{m} }{text{s}}$

$$
{color{#4257b2}v = lambdacdot f}=left( 0.6text{ m}right)cdotleft( 20text{ Hz}right) = 12text{ m/s}
$$

$$
text{color{#4257b2}12 m/s}
$$