To solve this problem we will apply the concepts related to the calculation of significant figures under tolerance levels. We will also take into account that the number of significant digits at the end of an answer must not be greater than the number of significant digits of a number:
PART A) The thickness of the 52 cards is

The thickness, t, of 1 card can be:


The thickness of 52 cards has 3 significant figures and the uncertainty has 1 significant digit. So
the significant figures of the thickness of one card and the uncertainty should also be 3 and 1 respectively


Therefore the thickness of one card is 
PART B) One card has uncertainty of 0.0001 in if measured using 1 deck.
The number of decks, n, required to create the uncertainty of 0.00002 in is


Therefore, 5 decks are required to measure the thickness of one card with an uncertainty of 0.00002 in