If A and C are independent, is P(A,C) = P(A)*P(C) always true? Here is the original question.

Of three possible events, event A is independent of the other two, and events B and C are mutually exclusive.  The probabilities that the individual events A, B, and C will occur are 0.5, 0.3, and 0.2, respectively.  What is the probability that both event A and event C will occur?

The answer to this question is:

Start with the “mutually exclusive” events, as this is the most restrictive statement.  If event B happens, event C cannot happen.  Likewise, if event C happens, event B cannot happen.  It is possible that neither event B or C will happen, but they can’t both happen.
Consider the possibilities, starting with whether event B happens.
If event B occurs, event C cannot occur, so there is no way for both event A and event C to happen. (i.e. Probability of both A and C is zero if B occurs.)  If event B does not occur, event C might happen, as might event A. 
Thus, the probability that both event A and event C will occur is the probability that B will NOT happen, A will happen, and C will happen.  {NOTE:  "and" means multiply probabilities, "or" would mean add probabilities.}
P(A and C) = P(not B) × P(A) × P(C)
  P(A and C) = [1 – 0.3] × 0.5 × 0.2 = 0.7 × 0.5 × 0.2 = 0.07 = 7%
The correct answer is 7%
Source: http://www.manhattanprep.com/gre/ChallengeProblems/LastWeek/

My question is:
1) P(A,C) = 0.07 in this question. However, this is not P(A)*P(C)=0.10, despite A and C are independent. Why does the rule P(A,C) = P(A)*P(C) fail even though A and C are independent? Is there a certain restraint that applies to this rule?
2) Event B and C are not independent. However, the problem states that P(A,C,~B) = P(A)*P(C)*P(~B). I thought this was possible only if C and ~B are independent. Can you please explain if this is valid?
Your help is greatly appreciated. Have a wonderful day.
 A: The stated answer of 0.07 is wrong. I will endeavor to show the mistake in the answer's logic.
They are attempting to use the rule of total probability to calculate $P(A\cap C)$. (The expression $A\cap C$ is like saying $A$ and $C$ happen simultaneously.) This procedure is as follows
$$P(A\cap C) = P(A\cap C\cap B^c) + P(A\cap C\cap B) = P(A\cap C\cap B^c).$$
(Noted that $B^c$ is the same as $\tilde{}B$.) We can eliminate $P(A\cap C\cap B)$ because $C$ and $B$ are mutually exclusive. As the OP noted, $C$ and $B^c$ are not independent so we cannot use the product rule. This is where the quoted solution goes wrong.
The correct answer can be gotten directly via the product $P(A)P(C)$ by using independence,  but I will also show how the calculation of $P(A\cap C\cap B^c)$ can be done using conditional probability. Using the definition of conditional probability, we have
$$P(A\cap C\cap B^c) = P(A | C\cap B^c)P(C\cap B^c) = P(A)P(C\cap B^c).$$
We were able to simplify $P(A | C\cap B^c)$ because $A$ is independent of $B$ and $C$. However, now we must calculate $P(C\cap B^c)$. We can rewrite this expression using the rule of total probability as
$$P(C\cap B^c) = P(C) - P(C\cap B) = P(C)$$
Because $B$ and $C$ are mutually exclusive, $P(C\cap B)=0$. And so we conclude that $P(A\cap C) = P(A)P(C)$ as given by the definition of independence.
For independent events, the calculation $P(A\cap C)=P(A)P(C)$ is always true. This result is given as a definition in most cases.
