There are 36 different rolls, which we can write as ordered pairs.

$$\Omega = \{ (x,y): x,y \in \{1,2,3,4,5,6\}\}$$

When the problem says dice, we understand that this means that each of the 36 pairs has the same probability, 1/36. We can count that there are 18 pairs in which the first die is odd, 18 pairs in which the second die is odd, and 18 pairs in which the sum is odd.

$P(A)=\frac{1}{2}$, $P(B)=\frac{1}{2}$. $P(C)=\frac{1}{2}$.

Also, $P(A\cap B)=P(\text{both are odd})=\frac{9}{36}=\frac{1}{4}$. We note that$P(A\cap C)=P(B\cap C)=\frac{1}{4}$.

Since $P(A\cap B) = 1/4 = P(A)P(B)$, events A and B are independent. By the same argument, B and C are independent and A and C are independent.

For A B and C to be mutually independent, we would need $P(A\cap B\cap C) = P(A)P(B)P(C)$.

However, the left hand side is $P(A\cap B\cap C) = P(\text{both are odd, but the sum isodd}) = P(\emptyset)=0$. By contrast, the right hand size is $(1/2)(1/2)(1/2) = 1/8$.

Since $P(A\cap B\cap C)\neq P(A)P(B)P(C)$, the events are not independent.