Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$ E_n = \left(n - \frac{1}{2}\right)\hbar \omega_e\tag {where $n=1,2,3,...$}$$
So, the energy levels of a harmonic oscillator are at $n=1,2$ and $3$.
$$ E_1 = \left(1 - \frac{1}{2}\right)\hbar \omega_e=\frac{1}{2} \hbar \omega_e$$
$$ E_2= \left(2 - \frac{1}{2}\right)\hbar \omega_e=\frac{3}{2} \hbar \omega_e$$
$$ E_3= \left(3 - \frac{1}{2}\right)\hbar \omega_e=\frac{5}{2} \hbar \omega_e$$
Now we need to find $\omega$ where we know that the angular frequency of a mass on a spring is given by $\omega=\sqrt{k/m}$.
So,
$$ E_1 = \frac{\hbar}{2} \sqrt{\dfrac{k}{m}}$$
Plug the known;
$$ E_1 = \frac{(1.05\times 10^{-34})}{2} \cdot \sqrt{\dfrac{(2)}{(9.11\times 10^{-31})}}$$
$$E_1=\bf 7.78 \times 10^{-20} \;\rm J=\color{red}{\bf 0.486 }\;\rm eV$$
By the same approach,
$$ E_2 = \frac{3(1.05\times 10^{-34})}{2} \cdot \sqrt{\dfrac{(2)}{(9.11\times 10^{-31})}}$$
$$E_2=\bf 2.33 \times 10^{-19} \;\rm J=\color{red}{\bf 1.46 }\;\rm eV$$
$$ E_3 = \frac{5(1.05\times 10^{-34})}{2} \cdot \sqrt{\dfrac{(2)}{(9.11\times 10^{-31})}}$$
$$E_3=\bf 3.89 \times 10^{-19} \;\rm J=\color{red}{\bf 2.43 }\;\rm eV$$
$$\color{blue}{\bf [b]}$$
To find the photon wavelength that is emitted when the electron undergoes $3\rightarrow 1$ quantum jump, we need to find the loss of energy which is given by
$$E_{\rm photon}=E_3-E_1=\dfrac{hc}{\lambda}$$
Solving for $\lambda$;
$$ \lambda =\dfrac{hc}{ E_3-E_1}$$
Plug the known and substitute the energies from above,
$$ \lambda =\dfrac{(6.63\times 10^{-34}) (3\times 10^8)}{ ( 3.89 \times 10^{-19} )-(7.78 \times 10^{-20} )}$$
$$ \lambda =\bf 6.39 \times 10^{-7} \;\rm m=\color{red}{\bf 639}\;\rm nm$$
