Step 0: Two identical packs together contain $2 \times 52 = 104$ cards.
Three cards are dropped, so the number of remaining cards $= 104 - 3 = 101$.

(i) Probability of a face card
Step 1: Face cards (King, Queen, Jack) in one pack $= 12$, so in two packs $= 24$.
Step 2: One face card (queen of hearts) was dropped, so face cards left $= 24 - 1 = 23$.
Step 3: $$P(\text{face card}) = \frac{23}{101}$$

(ii) Probability of a king or a queen
Step 4: Kings in two packs $= 8$ (none dropped). Queens in two packs $= 8$, one dropped, so $7$ left.
Step 5: Favourable outcomes $= 8 + 7 = 15$.
$$P(\text{king or queen}) = \frac{15}{101}$$

(iii) (a) Comparing the probability of a queen
Step 6: After dropping : $$P_1(\text{queen}) = \frac{7}{101} \approx 0{\cdot}0693$$
Step 7: If no card were dropped : $$P_2(\text{queen}) = \frac{8}{104} = \frac{1}{13} \approx 0{\cdot}0769$$
Step 8: Since $\dfrac{8}{104} > \dfrac{7}{101}$, YES, the probability of getting a queen would have been HIGHER if none of the cards were dropped, because one queen was among the dropped cards.

OR

(iii) (b) Probability of a jack
Step 9: No jack was dropped, so jacks left $= 8$ out of $101$ cards.
$$P_1(\text{jack}) = \frac{8}{101}$$
Step 10: If no card were dropped : $$P_2(\text{jack}) = \frac{8}{104} = \frac{1}{13}$$
Step 11: Since the numerators are equal and $101 \frac{8}{104}$$
Hence the probability of getting a jack is HIGHER in the case when the three cards were dropped.