2. no and yes (perl treats while1 as one identifier)

3a. yes

8a. yes, cool, unexpected solution :-)   our official version uses a much much more obscure trick.. funny we never saw this one
Try this instead: eval(q[$$foo]) ne eval(q["$$foo"])
•Update: actually, no, the original was fine too.. even though $$foo and $$foo + 0 print differently, they don't compare numerically unequal, which is what the exercise was

11. doesn't appear to work for me..

% perl -e 'sub STDOUT{\*STDOUT} <STDOUT> eq <+STDOUT> or die'
Died at -e line 1.
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12. yes, but why? :-)

•Update: fixed the "try this".. initially said "!=" instead of "ne"