Per OP's request.
First of all, as I understand the term orthogonal, as it is being used here, it is being used as a synonym for the phrase mutually exclusive.
Suppose that you have two events: $A$ and $B$.
To facilitate the analysis, make the simplifying assumption that
$p(A) \neq 0 \neq p(B).$
Then the two events either are or are not mutually exclusive.
Further, the two events either are or are not independent.
That gives 4 possibilities.
Case 1 $A$ and $B$ are both mutually exclusive and independent.
As has been discussed, this case is impossible, so no example can be provided.
Case 2 $A$ and $B$ are mutually exclusive but are not independent.
Example:
$A$ is the event of rolling a 1 on a die.
$B$ is the event of rolling a 6 on a die.
It is impossible for both events to simultaneously occur. Therefore, they are mutually exclusive.
$p(A) = (1/6)$ and $p(A|B) = 0$.
Since $p(A) \neq p(A|B)$, the two events are not independent.
Therefore, the two events are dependent.
Case 3 $A$ and $B$ are not mutually exclusive but are independent.
Example:
$A$ is the event of rolling a [1 or 2] on a die.
$B$ is the event of rolling [an even number] on a die.
It is definitely possible for both events to simultaneously occur. This is illustrated by rolling a [2]. Therefore, the two events are not mutually exclusive.
$p(A) = (1/3).$
Assume event $B$ occurred.
Then, either a [2, 4, or 6] was rolled.
Under this scenario, chance of $A$ occurring given that $B$ occurred is still (1/3).
Therefore, $p(A|B) = (1/3) = p(A).$
Therefore, these two events are independent.
Note that the exact same conclusion will inevitably be drawn if the primary focus here is event $B$, rather than event $A$.
$p(B) = (1/2).$
Assume event $A$ occurred.
Then, either a [1, or 2] was rolled.
Under this scenario, chance of $B$ occurring given that $A$ occurred is still (1/2).
Therefore, $p(B|A) = (1/2) = p(B).$
Therefore, these two events are independent.
Case 4 $A$ and $B$ are not mutually exclusive and are not independent.
Example:
$A$ is the event of rolling a [1 or 2] on a die.
$B$ is the event of rolling [1, 2, or 3] on a die.
It is definitely possible for both events to simultaneously occur. This is illustrated by rolling a [2]. Therefore, the two events are not mutually exclusive.
$p(A) = (1/3).$
Assume event $B$ occurred.
Then, either a [1, 2, 3] was rolled.
Under this scenario, chance of $A$ occurring given that $B$ occurred is (2/3).
Therefore, $p(A|B) = (2/3)$ and $p(A) = (1/3).$
Therefore, these two events are not independent.
Therefore, these two events are dependent.
Note that the exact same conclusion will inevitably be drawn if the primary focus here is event $B$, rather than event $A$.
$p(B) = (1/2).$
Assume event $A$ occurred.
Then, either a [1, or 2] was rolled.
Under this scenario, chance of $B$ occurring given that $A$ occurred is (1).
That is, if event $A$ occurred, then it is absolutely certain that event $B$ also occurred.
Therefore, $p(B|A) = (1)$ and $p(B) = (1/2).$
Therefore, these two events are not independent.
Therefore, these two events are dependent.
Best Answer
First question: Yes, the definition is correct. If $A$ and $B$ are mutually exclusive, then: $$P(A \cap B)=0$$ and: $$P(A \cup B) = P(A) + P(B)$$ and so indeed we have that: $$P(A \cup B \cup (A \cap B))=P(A) + P(B)$$
Second question: Your events are not mutually exclusive, and so you cannot add up the probabilities: you can only use $$P(A \cup B) = P(A) + P(B)$$ when $A$ and $B$ are mutually exclusive. And your events are not mutually exclusive because any two of them can happen in the same scenario. In fact, they can all three happen: you throw a tail for all three tosses.