Direction of motion in a magnetic field question

By exhaustion. It could be travelling vertically, but the emf wouldn't be Blv or B/lv, so it's not either of those two answers. What would the emf be if the wire was travelling upwards?

Well, I thought that since the wire would be cutting the flux at right angles, it would be, emf = (-) faraday's law

The magnetic force tries to push electrons through the wire, and this creates the EMF. If the wire was traveling upwards then the magnetic force would push electrons along the sides of the wire. However, the wire is considered to have negligible thickness.
You can see here the proof that $\dfrac{\mathrm{d}\Phi_B}{\mathrm{d}t}=Bul$. You do not need to consider the Lenz's law yet. You take this law into consideration when you have to decide about the polarity or the direction of any induced current.

I'm not completely sure what you mean by (-) Faraday's law. Do you mean$\dfrac{\mathrm{d}\Phi_B}{\mathrm{d}t},\ - \dfrac{\mathrm{d}\Phi_B}{\mathrm{d}t}$ or something else?
I think you're on the right lines. So what's $\dfrac{\mathrm{d}\Phi_B}{\mathrm{d}t}$?

Is it the case where any movement of the wire vertically or horizontally would induce an emf, but it's just the magnitude of the emf/equation you use differ?

The equation you always use is Faraday's law. Movement of the wire doesn't have to generate emf. For example if it moves vertically the emf is 0, since the width of the wire is 0, so there's no area, so no change in flux. Horizontal movement will create an emf, as long as the movement isn't parallel or antiparallel to the direction of the magnetic field.

Is it the case where any movement of the wire vertically or horizontally would induce an emf, but it's just the magnitude of the emf/equation you use differ?