I'm not too sure this will be the fastest way for your particular goal, but it's an interesting alternative route, so I'll present it anyway.

Make a copy of your first string, insert an extra "0" on the front, and do bytewise XOR. You'll end up with a string of consisting of chr(0) and chr(1), the latter if both "bits" were different. If you OR with a string of "0" characters, you'll turn that into "0" and "1" respectively. Like this:

my $str = "1010101011";
my $second = "0$str";
my $xor = $str ^ $second;
my $result = $xor | "0" x length $xor;
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After this, $result contains 11111111101. The first and last character are rather meaningless. But you can test whether bits with position 2, 5 and 8 pass, by extracting those characters from $result using substr.

Now, moving on to the second part of your question, You can use AND. First, make a mask, consisting of "0" characters and with the same length, setting it to "1" for those positions you're interested in. AND this with $result, and if this result is the same as $mask itself, you have a match.

my $mask = "0" x length $result;
substr($mask, $_, 1) = "1" foreach 2, 5, 8;
print "It passes!" if ($mask & $result) eq $mask;
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and it does pass.