Alright here goes. Admittedly, this game in my opinion was the more difficult of the four, still it wasn't bad. So five cities, unlimited number of flights, each one connected by the other...Hmm, this seems like a great time to use Manhattan's open board method. When problems match this method, open grouping, you should expect numerical distribution problems and some quick inferences.
H, M, P, T, V
M - only one
H --> ~T
H --> H + T
P + T --> P ~V
This game requires a couple quick inferences up front, and if I had to guess I'd say front-loaded game. First thing I did was see the most each could be with. I'm not going to include my diagram, mainly because I don't know exactly how to make it look effective. But I tried to describe it below.
So you know that H can only be with two cities at the most, since it can't be with M, because M can only have one and if H +M --> M +H +T, which violates a rule, and it can't be with T, per the rules.
Since H can't be with T, you can infer that T can only be with three cities at the most, M, T, and V. Furthermore, you can also infer that P can only connect with three cities at the most, because it cannot have H/T and V. Finally, V can be with four cities at the most, since there's not any real restrictions on it.
Moving into the questions which require more than a slight amount of thought, #12, 15, 16, and 17.
12. If three with P, you know they must be HMT, because P can only connect with HMTV, since it is with three regardless of if it chooses H, it must choose T, therefore, ~V.
~A, ~C, ~D, because P is with M, therefore nothing else is with M
~E, because P with HTM, not V
15. If T is only one with P, then M can't receive P or H, P can't receive H, according to the rules, T can't receive H, therefore V must receive H and T. So now you have H receiving only one, V, M could receive T or V, P only receives T as per the question, T receives at least V and P, and V receives at least H and T.
~A, because one must have H
~B, because V must have H, since H must have V, therefore V must have T as well
~C, H can only be with V
~E, T can't be with H
16. Max number of pairs. 13 open spaces, so initially cross out anything over six. Now consider the PT --> ~PV rule. You know H cannot receive T or M, so at the most it receives two, V and P, you know M receives one of P/T/V, P receives H, therefore receives T and can't receive V, T receives P, and since V receives H it must receive T, therefore T receives V. Now the question is which of P/T/V does M receive? So ignoring M, we have four pairs, this eliminates the rest, but since M is only with one of P/T/V, therefore there must be five pairs.
17. If four with the one remaining, then what must be true? Since V is the only one which receives four, then H, M, P, and T receive V. Since P + V, P cannot receive T, therefore it cannot receive H, and since it cannot receive M, because M receives only one which in this case is V, P is maxed out. Since T cannot receive H, P, since then P would receive T, which we know is false, or M, it must be maxed out. Since H can't receive P, T, or M, since they cannot receive H, H is maxed out. If you followed this thought process, the question is a cakewalk.