Beside the flush, the bigger pair doesn`t count here ?
Repeat after me: poker hands have five cards. EXACTLY five cards. No more, no fewer. In Hold'em, each player plays the best 5-card hand he can out of the seven available.
Vlad's best 5-card hand is A-8-7-4-2 of clubs. His opponent's best 5-card hand is A-8-7-4-2 of clubs.
If, perchance, our hero had, say, the 6 of clubs in his hand, then his best five-card hand would have been A-8-7-6-4 of clubs, for the win. But such is not the case.
No one can play a flush here without playing exactly the board.
The winner of a hand is defined by the best combination of 5 cards each user has. In this case the best combination each player has is a flush. The rest doesn't matter.
If you have a pair 2♠2♣ and the board would be 8♣8♠7♣7♠A♣ you would not have 3 pairs.
Because the best combination with 5 cards you can have in this case would still be the board 8♣8♠7♣7♠A♣.
Your example is the same both players have 7♣4♣2♣8♣A♣ as their best 5 card combination. So the rest of the combinations don't count anymore.