f(x)=4^x/(4^x +2)
=(4^x+2-2)/(4^x +2)
=1-2/(4^x +2)
故(1)f(a)+f(1-a)=1-2/(4^a +2)+1-2/[4^(1-a) +2]
=2-{2/(4^a +2)+2/[4/4^a+2]}
=2-{2/(4^a +2)+4^a/(2+4^a)}
=2-{(2+4^a)/(4^x +2)}
=1
(2)f(1/1001)+f(2/1001)+f(3/1001)+……+f(1000/1001)
=f(1/1001)+f(1000/1001))+f(2/1001)+f(999/1001)
+.+f(555/1001)+f(556/1001)
=[f(1/1001)+f(1-1/1001)]+[f(2/1001)+f(1-2/1001)]+.+[f(555/1001)+f(1-555/1001)]
=1+1+.+1(共計(jì)555個(gè)1)
=555