jQuery return ajax result into outside variable

You are missing a comma after

'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' }

Also, if you want return_first to hold the result of your anonymous function, you need to make a function call:

var return_first = function () {
    var tmp = null;
    $.ajax({
        'async': false,
        'type': "POST",
        'global': false,
        'dataType': 'html',
        'url': "ajax.php?first",
        'data': { 'request': "", 'target': 'arrange_url', 'method': 'method_target' },
        'success': function (data) {
            tmp = data;
        }
    });
    return tmp;
}();

Note () at the end.