A fair coin is flipped three times. What is the probability that tails occur exactly 1 time if it is known that tails occur at least once? ?

H: head. and P(H) = 1/2
T: tail, and P(T) = 1/2

P(at least 1T in 3 flips)
= 1 - P(3H in 3 flips)
= 1 - (1/2)³
= 7/8

P({exact 1T in 3 flips} and {at least 1T in 3 fligs})
= P(exact 1T in 3 flips)
= P(THH) + P(HTH) + P(HHT)
= (1/2)³ × 3
= 3/8

The required probability
= P({exact 1T in 3 flips} | {at least 1T in 3 fligs})
= P({exact 1T in 3 flips} and {at least 1T in 3 fligs}) / P(at least 1T in 3 flips)
= (3/8) / (7/8)
= 3/7

Flip a coin three times, what is the probability only one of the three is tails? (meaning the other two are heads...)

If you don't know where to start, you could list all possible outcomes for three flips:

Heads, Tails, Tails (HTT)
Heads, Heads, Tails (HHT)
Heads, Tails, Heads (HTH)
Tails, Heads, Heads (THH)
Tails, Tails, Heads (TTH)
Tails, Heads, Tails (THT)
Tails, Tails, Tails (TTT)
Heads, Heads, Heads (This one doesn't count since your problem tells you that tails will/must occur at least once)

So just check how many of those possibilities has only one Tails. I found three, out of 7 possibilities. So the probability is 3 outcomes with Tails occuring just once.. out of 7 possible outcomes. 3/7 

One flip got tails for sure, problem says.  That leaves two flips to have heads for sure or no win.  Each of the two flips has a fifty-fifty chance of heads or tails.  So for both the flips to be heads, that is 1/2 x 1/2 or a 1 in 4 or 25% chance the problem will be satisfied.