The first 7 games are finished, A leading 4 wins to 3. If A wins game 8 (50% chance), then A leads 5-3 and the match ends. So, half of the times, A wins.
In the remaining half of the times (i.e. if B wins game 8, for a score of 4-4), the game is even: either A or B will win game 9 and make the final point.
So, 1/2 of the times A wins, and in the remaining 1/2 of the times, A wins with probability 1/2 and so does B.
This means that A wins with probability
[tex]\dfrac{1}{2}+\dfrac{1}{2}\dfrac{1}{2}=\dfrac{3}{4}[/tex]
And B wins with probability
[tex]\dfrac{1}{2}\dfrac{1}{2}=\dfrac{1}{4}[/tex]
So, A should take 3/4 of the pot.