Solution:

a. No. Four apples with weights $1, 5, 7, 9$ and with weights $2, 4, 6, 10$ both give the results $6, 8, 10, 12, 14, 16$ when weighed in pairs.

b. Yes. Let $a \leq b \leq c \leq d \leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and dividing the sum by 4, we obtain $a + b + c + d + e$. Subtracting the smallest and the largest pairwise weights $a + b$ and $d + e$ from this we obtain $c$. Subtracting $c$ from the second largest pairwise weight $c + e$ we obtain $e$. Subtracting $e$ from the largest pairwise weight $d + e$ we obtain $d$. $a$ and $b$ are similarly determined.

c. Yes. Let $a \leq b \leq c \leq d \leq e \leq f$ be the weights of the apples. As each apple is weighed 5 times, by adding all 15 pairwise weights and dividing the sum by 5, we obtain $a + b + c + d + e + f$. Subtracting the smallest and the largest pairwise weights $a + b$ and $e + f$ from this we obtain $c + d$.
Subtracting the smallest and the second largest pairwise weights $a + b$ and $d + f$ from $a + b + c + d + e + f$ we obtain $c + e$. Similarly we obtain $b + d$. We use these to obtain $a + f$ and $b + e$.
Now $a + d, a + e, b + c$ are the three smallest among the remaining six pairwise weights. If we add these up, subtract the known weights $c + d$ and $b + e$ from the sum and divide the difference by 2, we obtain $a$. Then the rest follows.