OK so this is a tricky one, but good for the thinking brain.

So calling the resistance of each bulb R

and noting that the toal current always passes through R1

\(\displaystyle {I_{open}} = \frac{V}{{R + \frac{R}{2}}} = \frac{{2V}}{{2R + R}} = \frac{2}{3}\left( {\frac{V}{R}} \right)\)

\(\displaystyle {I_{closed}} = \frac{V}{{R + \frac{R}{3}}} = \frac{{3V}}{{3R + R}} = \frac{3}{4}\left( {\frac{V}{R}} \right)\)

\(\displaystyle {I_{closed}} > {I_{open}}\quad {\rm{since}}\quad \frac{3}{4} > \frac{2}{3}\)

Which immediately tells us the Q3 is true and Q4 is False since the greatest current passes when S is closed.

Each of the parallel bulbs carries the same current since they have equal resistance.

So when S is open 2 and 4 each carry half the total current or

\(\displaystyle \frac{1}{2}{I_{open}} = \frac{1}{2}\frac{2}{3}\left( {\frac{V}{R}} \right) = \frac{1}{3}\left( {\frac{V}{R}} \right)\)

And when S is closed 2, 3 and 4 carry one third of the total current

\(\displaystyle \frac{1}{3}{I_{closed}} = \frac{1}{3}\frac{3}{4}\left( {\frac{V}{R}} \right) = \frac{1}{4}\left( {\frac{V}{R}} \right)\)

So the parallel resistors carry more current when S is open rather than when it is closed

Immediately this answers Q1 and Q5 as True since one third is greater than one quarter.

(I note here that you have opposite answers for these, although they should be the same.)

Finally Q2 is False since only half the current through bulb 1 passes through bulb 2, the other half passing through bulb 4