4-1.a   One possible answer:
        Uniformly at random, pick n/2 bits to turn on.

4-1.b   One possible answer:
        The neighbors are all states that can be reached by turning a
        0 into 1 and a 1 into a 0.

4-1.c   Suppose two parents are 1010 and 0101. By uniform crossover,
        one possible result is the all zero and all one child.
        (Not balanced.)
        FYI, uniform crossover means for each bit position randomly
        pick a bit from the correspoinding bits of two parents.

4-1.d   One possible answer:
        We have two n-bit parents, A and B, and we want to create two
        n-bit children, C and D, by mixing the bits from A and B.
        In the bit positions in which A and B both have 1s or both
        have 0s, the children will get the same.  Now, note that if
        there are x bit positions in which parent A has a 
        1 and parent B has a zero, then there are also x bit positions
        with the reverse.  For each of the "0-1"s, we flip an unbiased
        coin to decide which child will get which bit.  After we're
        finished with the "0-1"s, say we sent y 1s to C and y 0s to
        D.  Now, pick a random set of y of the "1-0"s and send the 1s
        to D and 0s to C.
4-2 [ppt,pdf]