1:

$foo = *bar; \$foo =~ /^[A-G]/ and print "Ok!\n";
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7:

*foo = *1; !eval { ($foo) = $foo } and print "Ok\n";
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6: I don't know weather this is cheating or not, but it does make the (updated) expression true:

*foo = *_; sub { \$foo[0] == \$foo[1] && !$[ and print "Ok\n"; } -> ($
+x, $x);
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Update: a clean solution for 6:

sub{*foo=\@_}->($x,$x); \$foo[0] == \$foo[1] && !$[ and print "Ok!\n";
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Update again: 2:

*foo = \substr(1,0,1); \$foo =~ /^[H-N]/ and print "Ok!\n";
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Update: a wildcard solution for 3:

BEGIN{$^H{"qr"}=sub{""};$^H|=3<<16}; \$foo =~ /^\0/ and print "Ok!\n";
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Update: I finally realized that I should have known 12 from the start: it was in last year's quiz.

$[ = 2; *| = *[;  ($| = 1) == 2 and print "Ok!\n"; warn 0+$|;
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perl -wde 'sub DB::DB { ++$foo==2 and die "break out the eval"; }; !ev
+al { [ @foo ] } and print "Ok!\n"; '
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Note that I'm not just executing the print Ok statement with some trick. The expression !eval { [ @foo ] } actually gets evaluated, but somehow an exception gets generated inside the eval, thus eval returns false.

#!perl -d
sub DB::DB { ++$foo==2 and die "break out the eval"; }; !eval { [ @foo
+ ] } and print "Ok!\n";
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