Let the probability that the service life of components exceeds 800 hours be p (a). Because the service life of the three components obeys the normal distribution N( 1000, σ2),

The probability that the service life of each component exceeds 1200 hours is 13, so the probability that the service life of each component is less than 800 hours is 13.

Therefore, the probability that the service life of components 1, 2 and 3 exceeds 800 hours is 1- 13=23.

∴p(a)=( 1- 13× 13)×23= 1627，

So the answer is: 1627.