DFSA? NFSA? Regular Expression! (Part II)

So as we went over the basic on the previous post. I'll have something that's more challenging that'll take more effort to tackle it.

Give a DFA for each language below.

    (a) L_1 = { s (- {0,1}* : s contains at least 2 characters and s's
            second character is a 1 }

        We only need to keep track of which character we are processing for
        the first 2 characters, to ensure the second one is a 1.  Using the
        ASCII conventions from lecture (parentheses around accepting
        states, self-loops indicated by labels next to states, dead states
        not shown), we get the following DFA:

                           0,1        1
            --> q_0 ----> q_1 ----> (q_2) 0,1

        (Note: this has implicit dead state q_3.)

    (b) L_2 = { s (- {0,1}* : s contains fewer than 2 characters }

        Note that L_2 = { \epsilon, 0, 1 }.

                            0,1
            --> (q_0) ----> (q_1)

        (Note: this has implicit dead state q_2.)

    (c) L_4 = { s (- {a,b}* : every a in s is _eventually_ followed by b }
        For example, aaab (- L_4 because there is a 'b' that follows every
        'a'-- even though it is not immediately after the first two 'a's.

        Note that L_4 = { every string that does *not* end with an 'a' }.

                                 a
                        b      --->     a
            -->  (q_0)            ( q_1)
                               <---
                                  b