Help with Probability Question?

C: a person has Hepatitis C
U: IV Heroin user

P(a person is infected with Hepatitis C given this person is an IV Heroin user)
= P(C|U)
= 74%

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P(C|U) = P(C∩U) / P(U)
74% = P(C∩U) / 1%
P(C∩U) = 74% × 1%
P(C∩U) = 0.74%

P(a person infected with Hepatitis C is a IV Heroin user)
= P(U|C)
= P(C∩U) / P(C)
= 0.74% / 1.8%
≈ 41%

The answer to the first part (What is the probability that a randomly selected person is infected with Hepatitis C given this person is an IV Heroin user?) is already given in the question: 74% of IV Heroin users are infected with Hepatitis C

For the second part, this sort of question (conditional probability) is so easy if you just sketch a Venn diagram and use the ratio of the appropriate areas. It tends to get taught as Bayesian probabilities and all wrapped up in formulas and philosophy but actually it is just Venn diagrams and areas. 

What is the probability someone infected with Hepatitis C is a IV Heroin user? 

So, the whole world you are interested in reduces to people with Hep C (1.8%).
From the sketch, the absolute area inside our world that are users is 0.74 x 1%
And the probability is just the ratio of these areas,
p = 0.74 x 0.01 / 0.018
p = 0.41

You are told that 74% of heroin users are infected.  Therefore the probability that a user is infected, P(infected | user) = 0.74.

The second part is not exactly clear. Are the 74% of heroin users being counted as separate from the "general public", or are they included ?

Assuming that "general public" excludes the habitual heroin users, then
If 1% of the general public is a user, and 1.8% of those are infected,  P(gen.pub AND infected) = 0.01 * 0.018 = 0.00018

We already know that P(user AND infected) = 0.74

Therefore P(infected) = 0.74 + 0.00018 = 0.74018

Hence P(user | infected) = 0.74/0.74018 = 0.9996