Order statistics appear whenever a question involves the largest, smallest, or k-th value of a sample.
The technique for maxima
Always go through the CDF. The maximum is at most x exactly when every observation is at most x:
P(max <= x) = P(X1 <= x) times ... times P(Xn <= x)
For n independent uniforms on [0,1], that gives x^n, so the density is n x^(n-1) and:
E[max] = n/(n+1)
For minima, use the complement: P(min > x) = P(all > x). By symmetry, E[min] = 1/(n+1).
The spacings result
This is the elegant one and it comes up often.
Drop n uniform points on [0,1]. They divide it into n+1 intervals. Each interval has expected length exactly 1/(n+1).
So with 2 points, the three pieces average 1/3 each; the expected positions of the sorted points are 1/3 and 2/3. In general the k-th smallest of n uniforms has expectation k/(n+1).
The symmetry is the reason: all n+1 gaps are exchangeable, and they sum to 1, so each has expectation 1/(n+1). No integration needed.
Exponential minima
The other fact worth memorising. If X1, ..., Xn are independent exponentials with rates lambda_1, ..., lambda_n, then the minimum is exponential with rate equal to the sum of the rates.
The interview form: "three independent machines fail at rates ...; what is the expected time until the first failure?" Answer: 1 over the sum of the rates.
This follows immediately from the same complement trick - the minimum exceeds x only if all of them do, and the product of exponential survival functions is another exponential.
Common variants
Expected value of the second-largest of n uniforms? n/(n+1) minus the spacing, or directly (n-1+1)/(n+1) = n/(n+1) for the top and (n-1)/(n+1) for the second. Use the k-th smallest formula with k = n-1.
Expected range? E[max] - E[min] = (n-1)/(n+1).
Practise in random variables.