The longest piece

Idea. The maximum is awkward to handle directly, but its tail is easy: the longest piece exceeds $x$ if and only if some piece exceeds $x$. Inclusion–exclusion turns this into events about one, two or three pieces, and each of those has a simple probability.

Step 1: one piece. Call the lengths $L_1,L_2,L_3$ from left to right. The left piece is $L_1=\min (U,V)$, so $$\mathbb{P}(L_1>x)=\mathbb{P}(U>x,V>x)=(1-x{)}^2$$ for $0\le x\le 1$. The vector $(L_1,L_2,L_3)$ is uniform on the triangle $\{{\ell }_i\ge 0,{\ell }_1+{\ell }_2+{\ell }_3=1\}$, and this triangle is symmetric in the three coordinates. So every piece has the same law: $\mathbb{P}(L_i>x)=(1-x{)}^2$ for each $i$.

Step 2: several pieces. Asking that $k$ given pieces all exceed $x$ is the same as cutting a length $x$ off each of them and asking that what remains is a valid split. The remaining stick has length $1-kx$, and the uniform distribution on the triangle scales with it, so $$\mathbb{P}(L_i>x,L_j>x)=(1-2x{)}_{+}^2,$$ $$\mathbb{P}(L_1>x,L_2>x,L_3>x)=(1-3x{)}_{+}^2,$$ where $t_{+}=\max (t,0)$. Two pieces cannot both be longer than $1/2$, and three cannot all be longer than $1/3$, which is what the $(\cdot {)}_{+}$ records.

Step 3: the tail of the maximum. Let $M=\max (L_1,L_2,L_3)$. By inclusion–exclusion, $$\begin{array}{rl}\mathbb{P}(M>x)& =3(1-x{)}^2-3(1-2x{)}_{+}^2\\ & +(1-3x{)}_{+}^2.\end{array}$$ As a check, for $x<1/3$ this equals $1$ after expanding, as it should: some piece is always at least $1/3$.

Step 4: integrate. Using ${\int }_0^{1/k}(1-kx{)}^{2}dx=\frac{1}{3k}$, $$\begin{array}{rl}\mathbb{E}[M]& ={\int }_0^1\mathbb{P}(M>x)dx\\ & =3\cdot \frac{1}{3}-3\cdot \frac{1}{6}+\frac{1}{9}=\frac{11}{18}.\end{array}$$

So the longest piece has expected length $\frac{11}{18}\approx 0.611$.

Going further. The same argument with $n$ pieces gives $$\mathbb{E}[M_n]=\frac{1}{n}\left(1+\frac{1}{2}+\cdots +\frac{1}{n}\right).$$ For $n=3$ this is $\frac{1}{3}\cdot \frac{11}{6}=\frac{11}{18}$. The longest of $n$ pieces is about $\frac{\ln n}{n}$, only a factor $\ln n$ above the average length $\frac{1}{n}$.