Hey Keshav, I don't blame ya. This was definitely a tricky question and its asking for a lot. This question stumps a lot of students, actually, so thanks for asking this on the forum. I'm sure someone else will have a question about this too :)
To be able to complete this question, you need a strong understanding of meso compounds and optical activity, be willing to draw some things out to see it, and pay attention to when they are talking about 5 or 6 membered sugars. That's really all you need to know!
When we see step #1, We are told that the nitric acid reaction results in producing carboxylic acids on both ends of the molecule, and that glucose is still optically active. When you hear "optically active" you should be thinking of enantiomers and chiral centres, because this was where we encountered the concept. If we change both end groups into carboxylic acid groups, you will find that some of the molecules actually look like they are meso compounds with a mirror plane in the centre of the molecule. With this reaction, #1 and #7 are meso compounds. Since glucose is optically active after the reaction, #1 and #7 is ruled out.
When we see step #2, its essentially saying that we cleave off the top CHO group, and change the next carbon in the chain, as the new CHO group. This forms a 5-carbon chain sugar (which they also explicitly state again to be clear). The next step, is when this undergoes the reaction from step #1 (changing the ends to carboxylic acids), glucose is optically active. So from this step, #2, #5, #6 can be ruled out.
Step #3 is where it gets tricky and you really need to pay attention to the words here. The reaction is done from a pentose, meaning a 5-carbon sugar, to form glucose. It states, that when glucose AND its epimers are made from the pentamer and when done, and it undergoes the nitric acid reaction from step #1, we end up with optically active compounds. Since it states that it lengthes the aldehyde end to form a new aldehyde, and we are left with only #s 3, 4 and 8, we look at epimers of the first chiral carbon (from the aldehyde end), and note that if we were to put the first chiral hydroxyl group of #8 onto the right (therefore forming its epimer), we end up with NO optical activity since its a meso compound. Therefore, we can rule out #8.
Step #4 switches the CHO and CH2OH ends, and glucose results in a different molecule. You can try drawing this out and flipping it upside down. You will find that between whats left, #3 and #4, that we only make a new molecule with #3. And so therefore, we are able to identify that #3 is glucose.
Phew! That was long. I hope this helped. If you need more assistance, don't hesitate to reach out once more :)
Cheers, and happy studying!