Length contraction is given by 

$\displaystyle L_{rest}=\gamma L_{move}$

Where in this case,

$\displaystyle L_{rest}=5\textup{m}, \,\,\,L_{move}=3\textup{m}$

The Lorentz factor is given by:

$\displaystyle \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Combining these two equations, we get:

$\displaystyle L_{rest}=\frac{L_{move}}{\sqrt{1-\frac{v^2}{c^2}}}$

Solving for v:

$\displaystyle v=c\sqrt{1-\left ( \frac{L_{move}}{L_{rest}}\right )^2}= c\sqrt{1-\left ( \frac{3}{5}\right )^2}= c\sqrt{1-\left ( \frac{9}{25}\right )}$

$\displaystyle = c\sqrt{\left( \frac{16}{25}\right )}=\frac{4}{5}c=0.8c$

The total energy E of a relativistic particle is related to its rest mass energy E_o by:

$\displaystyle E=\gamma E_{o}$

Where gamma is related to the momentum by:

$\displaystyle p=\gamma mc$

Combining the equations and solving for p, we get:

$\displaystyle p=\frac{E}{E_{o}}mc=37 mc$

Which, in the units specified, is 37.

For a relativistic particle, momentum is given by:

$\displaystyle p=\gamma mc$

from which we can solve for gamma:

$\displaystyle \gamma=\frac{p}{mc}=\frac{50 mc}{mc}=50$

Total energy of a relativistic particle is given by:

$\displaystyle E=\gamma E_{o}=50E_{0}$

The question tells us:

$\displaystyle E=4E_{o}$

Where the rest mass energy of a particle is:

$\displaystyle E_{o}=mc^{2}$

Using the equation for the total energy of a particle, we can substitute:

$\displaystyle E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}$

$\displaystyle 4mc^2=\sqrt{p^2c^2+m^2c^4}$

Solving this for $\displaystyle p$, we find:

$\displaystyle p=\sqrt{15m^2c^2}=\sqrt{15}\,\,mc$

Which, in units of mc, gives us the correct answer.

The total energy of a relativistic particle is given by:

$\displaystyle E=\sqrt{p^2c^2+m^2c^4}$

Substituting the momentum, we get:

$\displaystyle E=\sqrt{9m^2c^4+m^2c^4}=\sqrt{10}mc^2$

Because the rest mass energy of a particle is given by:

$\displaystyle E_{o}=mc^2$

The total energy is:

$\displaystyle E=\sqrt{10}E_{o}$

The relativistic Doppler shift equation is given by:

$\displaystyle \lambda_{moving}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\lambda_{stationary}$

Where $\displaystyle \beta$ is defined as:

$\displaystyle \beta=\frac{v_{source}}{c}$

Because the stationary wavelength is shorter than the moving wavelength, the object must be receding from the Earth, eliminating two answers.

The speed of light is approximately $\displaystyle 3*10^8 \frac{m}{s}$, so $\displaystyle 3.2*10^8 \frac{m}{s}$ is not a possible answer.

Making the approximation that

$\displaystyle \frac{\lambda_{m}}{\lambda_{s}}=\frac{643.7}{212.5}\approx3$

Combining this with the first equation:

$\displaystyle \frac{\lambda_m}{\lambda_s}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}=3$

$\displaystyle 3\sqrt{1-\beta}=\sqrt{1+\beta}$

$\displaystyle 9(1-\beta)=1+\beta$

$\displaystyle 9-9\beta=1+\beta$

$\displaystyle 10\beta=8,\,\,\,\beta=0.8$

From beta, we can find the velocity:

$\displaystyle \beta=\frac{v_{source}}{c}\rightarrow v_{source}=\beta c=(0.8)(3\times10^8 \frac{m}{s})=2.4\times10^8 \frac{m}{s}$

The total energy of a relativistic particle is given by:

$\displaystyle E=\sqrt{p^2c^2+m^2c^4}$

Solving for the rest mass $\displaystyle m$:

$\displaystyle m=\frac{\sqrt{E^2-p^2c^2}}{c^2}=\frac{\sqrt{25-16}}{c^2}=\frac{3}{c^2}$

Which, in the units specified, gives the answer 3.

To apply a Lorentz transformation, we need gamma and beta:

$\displaystyle \beta=\frac{v}{c}=\frac{3}{5}$

$\displaystyle \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{5}{4}$

Then, apply the Lorentz transformation:

$\displaystyle x=x'=1$

$\displaystyle y=y'=0$

$\displaystyle z=\gamma(z'+\beta ct')=\frac{5}{4}(2+\frac{3}{5}\cdot 3)=4.75$

$\displaystyle ct=\gamma(ct'+\beta z')=\frac{5}{4}(3+\frac{3}{5}\cdot 2)=5.25$

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately $\displaystyle \frac{4}{3}$

$\displaystyle v'=\frac{c}{n}=\frac{3}{4}c$

To find the relative speed, use the following equation:

$\displaystyle v_{p}=\frac{v'+v}{1+\frac{vv'}{c^2}}$

$\displaystyle v\approx0.6c$

The Heisenberg Uncertainty principle states that

$\displaystyle \Delta x \Delta p \ge \frac{\hbar}{2}.$

Therefore, the smallest wavelength will have the lowest uncertainty in the position. Thus, we must pick the regime of options given which has the lowest wavelength. That is gamma-ray.