Prove difference between a set and the union of other two sets

Yes, your guesses are correct (except for “the necessary and sufficient condition for (i) to be correct” — I explain it in my other answer.)

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Ad (i):

To prove, that it is not always true, it is sufficient to provide particular sets $A$, $B$, $C$ which don't satisfy the equation (i.e. so called contra example).

So let $A$, $B$, $C$ are as in this table, where are progressively calculated expressions from them:

By comparing rows 5 and 7 we see that the equation is not satisfied.

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Ad (ii):

To prove the equality for every sets $A$, $B$, $C$, we need to prove that every element in the set on the left-hand side of the equality

$$A−(B∪C)=(A−B)−C$$

belongs to the right-hand set, and vice versa.

So let

$$x \in A−(B∪C).$$

Then by definition of the set subtraction

$$x \in A, \ \text{but}\ x \notin (B \cup C)$$

But if $x \notin (B \cup C)$, then by definition of the set union

$$\text{neither}\ x \in B, \ \text{nor}\ x \in C$$

which mean — again by definiton of the set subtraction — that

$$x \in (A - B), \ \text{ but }\ x \notin C$$

i.e. — once more by definiton of the set subtraction — that

$$x \in (A - B) - C$$

So we proved that the arbitrary element $x$ of the left-hand side belongs to the right-hand side, too.

The reversed statement we may prove by similar manner.