|PF 1|+|PF2|= 10

|F 1F2|=8

∫≈f 1pf 2 = 90

∴△F 1PF2 is a right triangle with a hypotenuse of |F 1F2|.

That is |PF 1|? +|PF2|？ =|F 1F2|？

∴(|PF 1|+|PF2|)？ -2 | pf 1 |×| PF2 | = | f 1 F2 |？

∴ 10? -4×( 1/2)| pf 1 |×| PF2 | = 8？

∴36-4S△F 1PF2=0

∴S△F 1PF2=9

Therefore, the area of Rt△F 1PF2 is 9.

(2)∵|PF 1|？ +|PF2|？ ≥2|PF 1|×|PF2|

∴|PF 1|？ +|PF2|？ +2 | pf 1 |×| PF2 |≥4 | pf 1 |×| PF2 |

∴(|PF 1|+|PF2|)？ ≥4|PF 1|×|PF2|

∴|pf 1|×|pf2|≤(|pf 1|+|pf2|)？ /4

From (1), we know | pf 1 |+pf2 | = 10.

Then |PF 1|×|PF2|≤25

∴ When |PF 1|=|PF2|, i.e. |PF 1|=|PF2|=5, "=" is obtained. At this time |PF 1|×|PF2| has a maximum value of 25.