Two functions with the same name in JavaScript - how can this work?

As far as I know, function foo() { aaa(); } is just var foo = function(){ aaa() } in JavaScript. So adding function foo() { bbb(); } should either overwrite the foo variable, or ignore the second definition - that's not the point. The point is that there should be one variable foo.

So, in this example the me variable should not be correctly resolved from inside the methods and it is not in Explorer 8 :-). I came to this example by trying to wrap them into another closure where (var) me would be, but I was surprised that it's not necessary:

    var foo = {
        bar1 : function me() {
            var index = 1;
            alert(me);
        },
        bar2 : function me() {
            var index = 2;
            alert(me);
        }    
    };

    foo.bar1(); // Shows the first one
    foo.bar2(); // Shows the second one

Demo: http://jsfiddle.net/W5dqy/5/

AFAIK function foo() { aaa(); } is just var foo = function(){ aaa() } in JavaScript.

Not quite; they're similar, but also quite different. JavaScript has two different but related things: Function declarations (your first example there), and function expressions (your second, which you then assign to a variable). They happen at different times in the parsing cycle and have different effects.

This is a function declaration:

function foo() {
    // ...
}

Function declarations are processed upon entry into the enclosing scope, before any step-by-step code is executed.

This is a function expression (specifically, an anonymous one):

var foo = function() {
    // ...
};

Function expressions are processed as part of the step-by-step code, at the point where they appear (just like any other expression).

Your quoted code is using a named function expression, which look like this:

var x = function foo() {
    // ...
};

(In your case it's within an object literal, so it's on the right-hand side of an : instead of an =, but it's still a named function expression.)

That's perfectly valid, ignoring implementation bugs (more in a moment). It creates a function with the name foo, doesn't put foo in the enclosing scope, and then assigns that function to the x variable (all of this happening when the expression is encountered in the step-by-step code). When I say it doesn't put foo in the enclosing scope, I mean exactly that:

var x = function foo() {
    alert(typeof foo); // alerts "function" (in compliant implementations)
};
alert(typeof foo);     // alerts "undefined" (in compliant implementations)

Note how that's different from the way function declarations work (where the function's name is added to the enclosing scope).

Named function expressions work on compliant implementations. Historically, there were bugs in implementations (early Safari, IE8 and earlier). Modern implementations get them right, including IE9 and up. (More here: Double take and here: Named function expressions demystified.)

So, in this example the me variable shoudl not be corectly resolved from inside the methods

Actually, it should be. A function's true name (the symbol between function and the opening parenthesis) is always in-scope within the function (whether the function is from a declaration or a named function expression).

NOTE: The below was written in 2011. With the advances in JavaScript since, I no longer feel the need to do things like the below unless I know I'm going to be dealing with IE8 (which is very rare these days).

var foo = (function(){
    var publicSymbols = {};

    publicSymbols.bar1 = bar1_me;
    function bar1_me() {
        var index = 1;
        alert(bar1_me);
    }

    publicSymbols.bar2 = bar2_me;
    function bar2_me() {
        var index = 2;
        alert(bar2_me);
    }

    return publicSymbols;
})();

var foo = (function(){
    return {
        bar1: bar1_me,
        bar2: bar2_me
    };

    function bar1_me() {
        var index = 1;
        alert(bar1_me);
    }

    function bar2_me() {
        var index = 2;
        alert(bar2_me);
    }
})();