A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, will the distance between them increase,decrease,or remain the same?

As the time increases the distance between the two drops
will also increase.  This is because both drops are subject to the acceleration of
gravity (9.8 m/s/s) and the first drop starts accelerating sooner so has reached a
higher velocity at each one second increment of
time.

Mathematically, the distance the drop has fallen is
determined by the equation:  delta y = v(i)t + 1/2 gt^2.  Where delta y is the distance,
v(i) is the initial velocity, g is acceleration of gravity, and t is the time.  Since
v(i) is zero, this reduces to:  delta y = 1/2gt^2.

Now
consider two drops able to fall a long distance.

at time
zero the first drop leaves the faucet and begins
accelerating.

at t = 1s, delta y = 4.9 m and the second
drop leaves the faucet.

at t = 2s, delta y = 19.6 m for
drop 1, 4.9 m for drop 2, a difference of 14.7 m.

at t = 3
s, delta y = 44.1 for drop 1, 19.6 m for drop 2, a difference of 24.5
m.

at t = 4 s, delta y = 78.4 m for drop 1, 44 m for drop
2, a difference of 34.4 m

at t = 5s, delta y = 122.5 m for
drop 1, 78.4 m for drop 2, a difference of 44.1 m.

So over
time the distance increases each second.