Questions

Yeah, I had a similar idea to Chase. Sort the targets in radial order (possibly using a custom comparator if you already have one?), then when looking for the closest for a specific target just check the ones on the left and the right to see which is closest. Use (index + list.size())%list.size() on the index for the left and right value so that it automatically wraps back to the beginning/end when you go over the edge. This approach would be the fastest, but I'm not sure it would be the smallest in terms of codesize.

Area Picture

Maybe this picture can explain what i need. As you can see the white lines are the absBearing to every enemy and the transparent white area is the area from the two enemys who have the widest bearing to each other. I need this for my radar and my movement.
And yes i need the biggest angle between two in radial order (i guess :)). Nevermind the blue and green lines they are for something else.

I'm not sure, that i correctly understood your question and that its algorithm is completely correct, but idea must work, i think

double minAbsBearing = Integer.MAX_VALUE;
double minRelBearing = Integer.MAX_VALUE;
double maxRelBearing = Integer.MIN_VALUE;

for (ATarget target : enemys) {
    double absBearing = ...; // me.angleTo(target) in my case
    double relBearing = Utils.normalRelativeAngle(absBearing);
    if (relBearing < minRelBearing) {
        minAbsBearing = absBearing;
        minRelBearing = relBearing;
    } 
    if (relBearing > maxRelBearing) {
        maxRelBearing = relBearing;
    }
}

double widestArea = maxRelBearing - minRelBearing;
double startAngle = minAbsBearing;
double endAngle = Utils.normalAbsoluteAngle(minAbsBearing + widestArea);

You nailed it mate. Thanks. Looks like i was totaly looking at the wrong direction. But at least my gut was right :). Man this is even smaller than i hoped for.

You are welcome, but this code would not work:) Now I have no time to describe why (writing on english is difficult for me), but try to test different cases and you will easy understand the problem.

I also thought there might be some issues at certain angles with the single loop approach.

I think you need to take a <math>n^2</math> approach, ie, a loop within a loop. Otherwise you need to use a sorting technique (with nlogn), then a single loop. Here is pseudocode:

initialise min_sum = Inf
For each bot, b:
    me_b = find the angle from me to b
    initialise min_rel = Inf, max_rel = -Inf.
    for each bot c, excluding b:
        me_c = find the angle from me to c
        rel = the relative angle between me_b and me_c
        if rel < min_rel, min_rel = rel. if rel > max_rel, max_rel = rel.
    sum = max_rel - min_rel
    if sum < min_sum, min_sum = sum, min_ang = me_b + min_rel, max_ang = me_b + max_rel.

The other approach would be to sort all the angles into radial order, then go through looking for the largest space between two angles, and then use the opposite part of the circle. So in pseudocode:

for each bot, b:
    me_b = angle from me to b
    angles[i++] = me_b

sort_ascending(angles)

max_rel = abs(relative_angle(angles[0] - angles[angles.length - 1]))
min_angle = angles[0]
max_angle = angles[angles.length-1]
max_index = 0
for(i = 1; i < angles.length; i++)
    rel = abs(relative_angles(angles[i-1] - angles[i]));
    if rel > max_rel
        max_rel = rel
        min_angle = angles[i]
        max_angle = angles[i-1]
        max_index = i