Part 1 is rather easy:

The chance to win at least one is determined as 1-(chance to win none).

Chance to win none is

995/1000*994/999*993/998*992/997*991/996*

990/995*989/994*988/993*987/992*986/991*

985/990*984/989*983/988*982/987*981/986

= 985*984*983*982*981 / (1000*999*998*997*996)

~ 0.9270

Chance to win at least one ticket is therefore 1-0.9270 = 0.072925 if each ticket is allowed to win (and thus every ticket you do not own can be a winner).

Chance to win at least one price is 7.29% with 5 tickets and 15 draws.

Working on part 2, which is a lot more tricky.

I forgot that trick, damn so easy xD

Isn't part 2 the same? 1000 tickets with 3.5 for each person. Making 1000/3.5 possible winners which is ~286.

Followed by the calculation:

__285 * 284 * 283 * 282 * 281 * 280 * 279 * 278 * 277 * 276 * 275 * 274 * 273 * 272 * 271__
286 * 285 * 284 * 283 * 282 * 281 * 280 * 279 * 278 * 277 * 276 * 275 * 274 * 273 * 272

= ~0.95

P = 1 - 0.95 = ~0.05

Or am I forgetting something?