Solution

a) In the moving frame there is no magnetic field, only an electric field. But in the Earth frame there are an electric field $\vec{E}$ and a magnetic field $\vec{B}$.
To prove the relation we will use the Lorentz transformation of the magnetic field to calculate the magnetic field in the moving frame; this is zero:

\begin{equation} \vec{B}' = 0 = \frac{\vec{B} - [\vec{\beta} \times \vec{E}]}{\sqrt{1-\beta^2}} \rightarrow \vec{B} = [\vec{\beta} \times \vec{E}] \end{equation}

In the equations above I set $c=1$ to simplify the calculations. Restoring the equation to include $c$:

\begin{equation} \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} \end{equation}

b) Using the previous equation we can calculate the magnetic field in each case:

\begin{equation} \vec{B} = \frac{[\vec{\beta} \times \vec{E}]}{c} \end{equation}

\begin{equation} \vec{B_1} = \frac{[\vec{\beta_1} \times \vec{E}]}{c} \end{equation}

\begin{equation} \vec{B_c} = \frac{[\hat{\beta} \times \vec{E}]}{c} \end{equation}

Where $\hat{\beta}$ is the unit vector in the direction of velocity. The magnetic field will be zero when the velocity is parallel to the electric field.