Average length of the longest segment
The answer to (B) is actually given in both Yuval Filmus' and my answers to the question about the average length of the shortest segment. It's $$\frac{1}{n} H_n,$$ where $H_n = \sum_{k=1}^n \frac{1}{k},$ i.e., the $n$th harmonic number.
"Clever" is of course subjective, but here's an argument for (A) in the $n$-piece case. At least there's only one (single-variable) integration in it. :)
If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$. So the $V_i$'s are the lengths of the pieces of rope.
The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$ This is proved formally in David and Nagaraja's Order Statistics, p. 135. Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$. For example, $P(V_1 > c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.
If $V_{(n)}$ denotes the largest piece of rope, then $$P(V_{(n)} > x) = P(V_1 > x \text{ or } V_2 > x \text{ or } \cdots \text{ or } V_n > x).$$ This calls for the principle of inclusion/exclusion. Thus we have, using the "key idea" above, $$P(V_{(n)} > x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$ where the sum continues until $kx > 1$.
Therefore, $$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} > x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk} $$ $$= \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$ where the last step applies a known binomial sum identity.
Neat as it is, I don't think Rahul's answer can be correct. If we have $3x+2y+z=1$ and $3x+2y+z=9n$, then $n=1/9$, which means $x \leq 2/9$, which can't be right, as one solution is all pieces being of length 1/3 (however unlikely this exact solution may be, $x$ can take values in $(2/9,1/3]$).
Stefan's answer is wrong because the probability of any given point on the stick being in the longest piece (when cut in 2) is not uniform. That can be seen by considering the point halfway along, which is always in the longest piece. You can find the probability density function for the likelihood of any given point being in the longer half, then integrate to get the answer below.
My preferred solution is to let the cuts be at $X, Y$, with $Y \gt X$:
Image of cut positions
Then each piece is equally likely to be the longest, and the expected length of the longest piece doesn't depend on which piece we choose. Then we can calculate $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$.
We have the three inequalities: $$X \gt Y-X \implies Y < 2X$$ $$X \gt 1-Y \implies Y > 1-X$$ and, from our setup, $$Y \gt X$$ These can be represented by the following diagram:
Diagram of inequalities
Then the area satisfying our inequalities is the two triangles A and B. So we wish to find the expected value of $X$ within this area.
The expected value of $X$ in A is $\bar{X}_A = \frac{1}{2}-\frac{1}{3}(\frac{1}{2}-\frac{1}{3}) = \frac{8}{18}$.
The expected value of $X$ in B is $\bar{X}_B = \frac{1}{2}+\frac{1}{3}(\frac{1}{2}) = \frac{4}{6} = \frac{12}{18}$
The area of A is $A_A = \frac{1}{2} \times \frac{1}{2}\times (\frac{1}{2}-\frac{1}{3}) = \frac{1}{24}$.
The area of B is $A_B = \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{8} = \frac{3}{24} = 3 A_A$.
So $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} ) = \frac{\tfrac{8}{18} + 3\left(\tfrac{12}{18}\right)}{4} = \frac{11}{18}$
(Apologies for the comments on other solutions and the links, but I haven't got a high enough reputation to comment or embed directly)