Partial solution
In general:
it seems that many of the bullet points (all?) have concealed numbers somehow. Perhaps five digits each.
Crazy robots: we want you to dance for Miss Evan.
there are a lot of bits of this that sound somewhat like digits: wan / t you / to / dance / for / mi / ssevan -> 1 / 2 / 2 / ? / 4 / ? / 7, perhaps. Levieux observes that craZYRObots is probably meant to yield a 0, and suggests ignoring the TYOU because it's not a terribly convincing 2 (thus yielding 5 digits 01247). That's probably right but I slightly wonder whether actually we should be finding 6 digits in each line to give a 6x6 square, in which case this would be 012247.
How I want a drink,” thought Akira, getting himself primed.
"How I want a drink" is the beginning of a famous mnemonic for the digits of pi. Count the letters: 3,1,4,1,5. "Primed" is also suggestive of mathematics; it might refer to prime numbers or to mathematicians' habit of denoting various things related to $x$ as $x'$ (sometimes read as "$x$ prime"). I don't know of any really eminent mathematicians called Akira, but ... aha! How about Akira Haraguchi, who has memorized startlingly many digits of pi? If this is just about finding five digits per line, clearly they are 31415; if not, perhaps it resolves to "pi" or something of the sort. ... BUT further developments -- see kayzeroshort's answer and comments thereto -- suggest that actually the digits we want here may be 83431, the number of digits of pi Haraguchi recited in his previous demonstration in July 2005.
You’re high, mom- they don’t have TV on the moon.
Still mysterious so far.
Oh mercy, my darling cokehead’s latest x-ray image is ideal!
Levieux astutely observes in comments that after the first word all the others begin with Roman numerals, so (since that first word is "Oh") perhaps this line leads to the number 02663.
Oscar dashed off a message: “Eight? Noooo. Three. Angst. Oaths.”
In comments, Levieux suggests taking those five words as Morse, with vowels and consonants indicating dits and dahs, giving us five digits again: 71367. Or, as Carmeister points out in comments, switching roles it could be 26812.
Pick a tether, you dick. No, not that tether. Jesus.
"Tether" and "dick" are numerals in traditional northern English sheep-counting systems. Yan, tan, tether, mether, pip, azer, sezar, akker, conter, dick. Oh, and apparently "pick" is 5 in some systems (thanks to Levieux and Neil W in comments for noticing this, which fixes the fact that I had some things backwards). If we're looking for five digits per line, we have 'em: 53103.
If I'm right about
each bullet point leading to some digits
and if
as seems likely (and explicitly conjectured by Levieux in comments) each leads to exactly five digits
then here is what we seem to have:
01247
31415
?????
02663
71367 or 26812
53103
What should we do with these having got them? Here are a few ideas. There are surely many more possibilities.
Read them (in some order) as decimal letter values, combining adjacent ones where appropriate. I've played with this just a little without success but have by no means tried everything that could be tried. Turn each digit into a single bit and read rows via the Baconian cipher, yielding a 6-letter word. Using parity to get those bits doesn't seem to work well. I haven't looked at other possibilities such as <5/>=5. Turn the digits into Morse (like in the 5th bullet point) and re-parse to get letters. I think this is unlikely to work -- Morse digits have too many repeated symbols. Turn each 5-digit number somehow into a 5-character imgur image name and look 'em up. 5 caseful alphanumerics yields a much larger space than 5 digits. I don't think this is promising.