19. Octave, 144 bytes

Most requirements up until now can easily be circumvented using comments and suppressed strings, making it trivial to add answers. I figured I'd make it a bit harder by disallowing some characters instead.

.6;%+->?|"e"|?>-+%;6.
f=@(x)all(ismember('Hi, Retina!',x))&sum(ismember(x,cat(2,33:36,46,91:93,'')))<6&x(2)<60
%

Try it online!

Satisfies:

2. The first character is a ..
3. It contains an e.
4. Its length is even.
5. Its length is a perfect square.
6. It contains an a.
7. It contains a > character.
8. Contains the exact string ->.
9. Contains the exact string Hi, Retina!.
10. The sum of the first two Unicode code points is a multiple of 5.
11. The 10-th character is a ".
12. The last non-empty line does not have any duplicate characters.
13. The first line is a palindrome of length > 5.
14. The first line is exactly 21 characters long (not including newline).
15. It contains a ?.
16. It contains a |.
17. Contains a +.
18. It is at least 28 lines long.
19. The following characters are used five times in total: !"#$.[\] and the codepoint of the second character is less than 60.

For future answers:

• The first character is a ..
• Its length is an even perfect square.
• Contains the exact sequence ->.
• Contains the exact string Hi, Retina!.
• The second character's Unicode code point, mod 5, is 4, and its code point is lower than 60.
• The 10-th character is a ", and so is the twelfth character (palindromic rule).
• The last non-empty line does not have any duplicate characters.
• The first line is a palindrome of length = 21
• It contains a ?.
• It contains a |.
• It contains a +.
• It is at least 28 lines long.
• The following characters can only be used five times in total: !"#$.[\].
• Note that four of those are in the first line, due to the palindromic rule, and the exclamation mark is in the 'Hi, Retina!' string.