is correct since the series studied in -spectrum, including Balmer series, are de-excitation series or emission series. So, electrons get de-excited to which means that .
It is possible to obtain I.E. from the formula above, but since the question has stated the formula for the Balmer series, nlower has been fixed as 2 . So, it is not possible to calculate I.E. from it. To calculate I.E., we'll have to put , which isn't possible here.

With a, increases as is increased. So, the last line of the Balmer series, i.e. from infinity to , will have the maximum energy in the series and thus, the lowest wavelength. Similarly, the first line in the series, i.e. from to will have the lowest energy in the series and thus, the highest wavelength. Which makes this statement correct.
As orbits with higher orbit number or those that are further away from the nucleus are considered, the energy gap in-between subsequent orbits decreases. Now, consider the following for example and with nlower fixed as 2 .

Energy of a photon released on transition from to will have similar energy to that of the photon that gets released on transition from to , because energy of the and the orbit will be very close in value. That means they will also have very close values of wavelengths, which further implies that these two lines will be situated quite close to each other on the photographic plate.

In a similar fashion, we can see that as the increases, the lines start to converge together. And since, increasing the will indeed lead to an increase in the energy of the photon released, it will end up releasing photons of shorter wavelengths. Combining these two statements we can easily see that as the wavelength decreases, the spectral lines start to converge.

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