A reference to a variable or array/key can only happen if the source exists. If it does not then what would the reference be to? However if $a[b] does not exist then what would happen to $c? What would it be a reference to?
So PHP must create $a[b] automatically. Doing so also requires creating the $a array, since that doesn't exist yet either, but just like during $a[b]=123 PHP does not warn when it happens automatically.

So
1a. There is no warning for implicitly creating arrays. This is not related to references.
1b. There is no warning when $a[b] was implicitly created because you did, after all, want a reference to it.
2. Since $a[b] had to be created, count($a) will return 1.

This is also mentioned in the docs.
http://php.net/manual/en/language.references.whatdo.php
> Note: If you assign, pass, or return an undefined variable by reference, it will get created.

This is a strange behavior.
If I do $c = $a['b']; I get a notice for trying to get the value of an undefined variable. But $c =& $a['b']; silently creates the variable.
I don't mind the array being created implicitly, but a notice should be generated since I am referencing something that does not exist.
Trying to get the value of something undefined: notice.
Trying to get the reference of something undefined: perfectly fine.

yes because otherwise functions like http://php.net/manual/en/function.exec.php would not be possible (or do you create $output and $return_var in your code  before? hint: strip that useless lines..)