The reason this smelled so of homework is that many instructors love these tricky regex puzzles. "Do <whatever> with a single regex" whether or not that is a sane thing to do. Most such problems are better solved with several regexen or other constructs, all pasted together with a bit of logic.

Your problems each consist of two distinct logical requirements.

1. An alphabet, and
2. A condition on the number of certain substrings.

The first is most easily handled with a negative binding of a negated character class.

sub is_abc {
    local $_ = shift || $_;
    $_ !~ /[^abc]/
}

sub is_ab {
    local $_ = shift || $_;
    $_ !~ /[^ab]/
}
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Those subs could easily be autoloaded.

The arithmetic tests differ, and their best implementation may or may not involve a regex match. The odd number of ab pairs is easily implemented with a global one, together with bitwise &,

sub has_odd_ab {
    local $_ = shift || $_;
    my $n = () = /ab/g;    # count matches
    1 & $n                 # odd?
}
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The condition for the second problem is better done with tr///, because it is quicker for counting individual characters. However it looks, tr/// is not really a regex solution. The logical or appearing in the condition allows a tidy way to package those conditions.

sub odd_a {
    local $_ = shift || $_;
    1 & tr/a//;
}
sub odd_b {
    local $_ = shift || $_;
    1 & tr/b//;
}
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and the solutions,

sub prob_one {
    local $_ = shift || $_;
    is_abc and has_odd_ab;
}
sub prob_two {
    local $_ = shift || $_;
    is_ab and ! odd_a or odd_b;
}
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All that localizing of $_ is to allow $_ to be the default argument without causing any modifications to it.