对于事件 $A$ 和 $B$，已知 $\mathrm{P}(A') = 0.3$ 且 $\mathrm{P}(B) = 0.55$。

1. Find the probability of event $A$.

$\mathrm{P}(A) = 1 - \mathrm{P}(A') = 1 - 0.3 = 0.7$

2. Use the definition of independent events to find $\mathrm{P}(A \cap B)$.

$\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.7 \times 0.55 = 0.385$

1. Determine the general upper and lower bounds for $\mathrm{P}(A \cup B)$.

$\mathrm{P}(A) = 1 - \mathrm{P}(A') = 1 - 0.3 = 0.7$

The maximum possible probability for any event is $1$, so $\mathrm{P}(A \cup B) \le 1$.

The union of two events must be at least as probable as each individual event:
$\mathrm{P}(A \cup B) \ge \max(\mathrm{P}(A), \mathrm{P}(B))$
$\mathrm{P}(A \cup B) \ge \max(0.7, 0.55) = 0.7$

So the general bounds are $0.7 \le \mathrm{P}(A \cup B) \le 1$.

2. Calculate the value of $\mathrm{P}(A \cup B)$ if the events were independent, and exclude it from the range.

If $A$ and $B$ were independent, then from part (i), $\mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) = 0.7 \times 0.55 = 0.385$.

Using the addition rule:
$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$
$\mathrm{P}(A \cup B) = 0.7 + 0.55 - 0.385 = 0.865$

Since it is given that $A$ and $B$ are NOT independent, this specific probability must be excluded.
Therefore, $0.7 \le \mathrm{P}(A \cup B) \le 1$, where $\mathrm{P}(A \cup B) \neq 0.865$.