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.

This is all you need to do:

var myVariable;

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

NOTE: Use of "async" has been depreciated. See https://xhr.spec.whatwg.org/.