You could also avoid going numeric altogether by converting your string to a bitstring (a string of ASCII 0's and 1's, that is, here):

my $testString = "          00008008 000000FF 00800000";
$testString =~ s/\s+//g;

# convert to a string representation of '0's and '1's
my $bits = unpack("B*", pack("H*", $testString));

# test if specific bit is set
my $maskResult = substr($bits, 1, 1) eq '1';

printf "%s\n--> 2nd bit from left: %d\n", $bits, $maskResult;
[download]

Checking for several non-adjacent bits with substr might get a bit unwieldy (if you need that) ... in which case you could use the bitwise string-AND operation. For example

my $testString = "          01008008 000000DF 00800000";
my $mask       =           "42000000 00000020 00000000";

for ($testString, $mask) {
    s/\s+//g;
    $_ = pack("H*", $_);  # convert from hex to (binary) bitstring
}

# bitwise string AND
my $maskResult = $testString & $mask;

# ----- visualise what's going on (functionally not required) -----
for my $bits ($testString, $mask, '', $maskResult) {
    if ($bits) {
        print unpack("B*", $bits), "\n";
    } else {
        print "-" x 96, "\n";
    }
} # ---------------------------------------------------------------

# check for any non-zero bytes
$maskResult =~ tr/\0//d;

printf "There were %s matching bits.\n", length($maskResult) ? "some":
+"no";
[download]

This would test whether any of the bits specified in $mask are set in $testString.

There is always more than one way to do it :)