A small amount of counting is used in (a). For (c), there is an easy no counting solution.
(a) We want the probability that the first four customers order a total of $2$ white and $2$ brown and the $5$th customer orders white. The probability of $2$ white and $2$ brown in $4$ orders is $\dbinom{4}{2}\left(\dfrac{1}{2}\right)^4$. Given that this has happened, the probability the $5$th order is white is $\dfrac{1}{2}$. For the required probability, multiply. We get $\binom{4}{2}\left(\frac{1}{2}\right)^5.$
(b) This is just the probability of three heads in a row when tossing a fair coin.
(c) This can be computed in various ways. One reasonable approach is by recycling. Let $x$ be the probability all the whites are ordered before any brown. Let $y$ be the probability the $3$rd white is the $4$th item ordered. And let $z$ be the probability that the $3$rd white is the $5$th item ordered. We want $x+y+z$. We already computed $x$ in part (b), and $z$ in part (a). It remains to compute $y$, which can be done in a way quite similar to the solution of (a).
But there is a trivial way of finding the answer! By symmetry, "we run out of whites before we run out of browns" is just as likely as "we run out of browns before we run out of whites." Thus the required probability is $\dfrac{1}{2}$.
Remark: The $\binom{4}{2}\left(\frac{1}{2}\right)^4$ that formed part of the answer to (a) did use counting. We need to know how many $4$-letter words there are that have $2$ W and $2$ B.