A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most two tails, then P(AUB) is:

Total outcomes when a fair coin is tossed 3 times:

\[ n(S) = 2^3 = 8 \]

Event \(A\): exactly 2 heads = {HHT, HTH, THH}

\[ |A| = 3 \]

Event \(B\): at most 2 tails = all outcomes except {TTT}

\[ |B| = 7 \]

Since every outcome of \(A\) (two heads ⇒ one tail) is included in \(B\), we have:

\[ A \subseteq B \;\;\Rightarrow\;\; A \cup B = B \]

Therefore:

\[ P(A \cup B) = P(B) = \frac{|B|}{8} = \frac{7}{8} \]

Final Answer:

\[ \boxed{\tfrac{7}{8}} \]