Empirical formula, simplifying problems
You find #2 difficult because #1 and #3 are ridiculously easy!

In these two questions the mass of every element present just happens to match exactly with the atomic weight of that element.  So when you divide each mass by the corresponding atomic weight, the resulting ratios are very close to integers, or integers divided by 100.  So the (almost) final step of dividing each ratio by the smallest ratio goes by almost unnoticed.

But consider for the moment what would happen if you used three or four decimal place atomic weights.  All those nice neat exact answers would disappear in a haze of "sort of" whole numbers, and you would have to use some judgement as to whether a number is close enough to a whole number.

In #2, the questioner made the "drastic" change of using 4.0 and 40.1 as the two numbers for an element.  If you had carried on mechanically, dividing all ratios by the smallest ratio, you would have gotten:


Ca ... $1.0$
O .....$2.005$
H......$2.005$


Is $2.005$ close enough to a whole number?  Or should you multiply everything by $1000$ to give $1000, 2005, 2005$ as the empirical formula?