Well, okay, so here's what I had in terms of proof for both problems:

I guess I should post my solutions now. Maybe I did make a mistake somewhere, there's people here better at this kind of maths than I am, so whatever.

Easy version: 2, 4, 4, 6 is the list, so n = 4. Mode is 4 (two fours), Median is 4 (4 and 4 in the middle), Mean is 4 (sum is 16, 16/4=4) and range is 4 (6-2 = 4). This is also unique as the solution - you need two or more fours. for the mode. 3-4 fours doesn't work, as then the range or mean will fail. So two fours for the mode. As the median is 4 they have to be the middle numbers. And as the range and mean are also 4, the last two numbers must be equidistant from 4, i.e. one is two less, one or two more.

You can't do it in three numbers, since getting a mode of 3 means two 3s, so for the range your final number must be 0 or 6, which makes your mean 0 or 6. You can't do it in two numbers, as to get a range of 2 and a mean of 2 your numbers must be 1 and 3, which has no mode. One number is has range 0.

Hard version: 6 numbers. Here's an example list: 2.5, 6.5 - sqrt(34)/2, 6, 6, 6.5 + sqrt(34)/2, 8.5.

1-4 numbers can't work, which follows for the working in the easy version. For 4 numbers, there's a unique layout, right? But that doesn't have variance 4 (it has a variance of 8/3). What about 5? Well, we need 5, 5 in there for the median. Let x be the smallest number in the list. Then x + 5 is the largest number (range), and finally 10 - 2x is the final number (mean). Then firstly 4/3 < x < 10/3 in order to ensure 10 - 2x is not outside our range. But then we run into issues. Plugging x, x + 5, 10 - 2x, 5, 5 into our variance equation - E((X-µ)^{2}) gives us:

25 = (x - 5)^{2} + x^{2} + (5 - 2x)^{2}

25 = x^{2} - 10x + 25 + x^{2} + 25 - 20x + 4x^{2}

25 = 6x^{2} - 30x + 50

0 = 6x^{2} - 30x + 25

x ~= 1.057 or x ~= 3.943

But both of these are outside our range of x, since they make the range of the list greater than 5. Therefore, 5 doesn't work.

So now 6. Well, we have a solution, so we're done! Let's find the family of solutions though. We need to have 6, 6 in there because mode. The smallest number we can call x again, and the biggest x + 6 (range). Let's call the 5th number y and then finally the 6th number will be 18 - 2x - y (mean). WLOG let y<=6 (median requires either it or 18 - 2x - y is less than or equal to 6 and the other greater than or equal).

Now we plug these into our variance formula and get:

36 = (6-x)^{2} + (6-y)^{2} + (6-6)^{2} + (6-6)^{2} + (2x + y - 12)^{2} + (6 - 6 + x)^{2}

Expanding and simplifying that gives us:

0 = 6x^{2} + 2y^{2} + 4xy - 60x - 36y + 180

I can't be bothered to rearrange this myself, but it looks like an ellipse or something. Let's have WA rearrange for y.

y = -sqrt(-2x^2 + 12x - 9) - x + 9

Great. So now our list of six numbers is: x, 9 - x - sqrt(-2x^2 + 12x - 9), 6, 6, 9 - x + sqrt(-2x^2 + 12x - 9), x + 6

And here's where I realised the mistake. So... The issue is that, when we start putting bounds on x, we find there's no value which works. It's hard to state exactly since there's a lot of little cases, but basically: Increase x and the 2nd term drops below 3, so we can't do that. However the second term does eventually start increasing, but not enough to overtake x before the 5th term drops below 6. So we can't increase x. Decreasing x causes the same issue, but in reverse. The 5th term starts increasing above x+6 It starts decreasing eventually but not enough before the 2nd term increases above 6. So there's no value of x which satisfies all needed properties, meaning 6 doesn't work.

So... yeah I'm slightly bummed now. Turns out (unsurprisingly) that florrat was correct.