Last Friday of each month

sub MAIN (Int $year = Date.today.year) {
    my @fri;
    for Date.new("$year-01-01") .. Date.new("$year-12-31") {
        @fri[.month] = .Str if .day-of-week == 5;
    }
    .say for @fri[1..12];
}

Example:

$ ./lastfri 2038
2038-01-29
2038-02-26
2038-03-26
2038-04-30
2038-05-28
2038-06-25
2038-07-30
2038-08-27
2038-09-24
2038-10-29
2038-11-26
2038-12-31

A solution without a result array to store things in:

sub MAIN (Int $year = Date.today.year) {
    say ~.value.reverse.first: *.day-of-week == 5
        for classify *.month, Date.new("$year-01-01") .. Date.new("$year-12-31");
}

Here, classify sorts the dates into one bin per month (but preserves the order in each bin). We then take the list inside each bin (.value) and find the last (.reverse.first) date which is a Friday.

Another variation where the data flow can be read left to right using feed operators:

sub MAIN (Int $year = Date.today.year) {
    .say for Date.new("$year-01-01") .. Date.new("$year-12-31") ==> classify *.month ==>
             map *.value.reverse.first: *.day-of-week == 5
}