Let $\mathcal{U}_n=\{U_1, \ldots, U_n\}$ be a collection of iid random variables. Denote $U_{1:n} = \min \mathcal{U}_n$ and $U_{n:n} = \max \mathcal{U}_n$ be the smallest and the largest of the collection.
Then
$$\begin{eqnarray}
F_{U_{1:n},U_{n:n}}\left(u_1, u_n\right) &=& \Pr\left(U_{1:n} \leqslant u_1, U_{n:n} \leqslant u_n\right) \\&=& \Pr\left(U_{n:n} \leqslant u_n\right) - \Pr\left(U_{1:n} > u_1, U_{n:n} \leqslant u_n\right)
\end{eqnarray}
$$
Now, since
$$
\{U_{n:n} \leqslant u_n\} = \{\max \mathcal{U}_n \leqslant u_n\} \equiv \{ U_1 \leqslant u_n, U_2 \leqslant u_n, \ldots, U_{n} \leqslant u_n \}
$$
and
$$
\{U_{1:n} > u_1\} \equiv \{ U_1 > u_1, U_2 > u_1, \ldots, U_n > u_1 \}
$$
We have,
$$
F_{U_{1:n},U_{n:n}}\left(u_1, u_n\right) = F_U\left(u_n\right)^n - \left(F_U\left(\max(u_n,u_1)\right)-F_U\left(\min(u_1,u_n)\right))\right)^n
$$
Differentiating, and assuming $0 < u_1 < u_n <1$:
$$
f_{U_{1:n},U_{n:n}}\left(u_1, u_n\right) = n(n-1) \left(F_U\left(u_n\right)-F_U\left(u_1\right))\right)^{n-1} f_U(u_n) f_U(u_1) = n (n-1) \left(u_n-u_1\right)^{n-2}
$$
The expected value of $R=U_{n:n}-U_{1:n}$
$$
\mathbb{E}\left(R\right) = \int_{0}^{1} \mathrm{d}u_n \int_0^{u_n} \mathrm{d}u_1 n \left(n-1\right) \left(u_n-u_1\right)^{n-1} = \frac{n-1}{n+1}
$$
Similarly we can compute $\Pr\left(R > \frac{1}{2}\right)$:
$$\begin{eqnarray}
\Pr(R > 1/2) &=& \Pr\left(u_1 < u_n - \frac{1}{2} \right) \\ &=& \int_{1/2}^{1} \mathrm{d}u_n \int_0^{u_n-1/2} \mathrm{d}u_1 n \left(n-1\right) \left(u_n-u_1\right)^{n-1} \\ &=& \int_{1/2}^{1} \mathrm{d}u_n n \left(u_n^{n-1} - \left(\frac{1}{2}\right)^{n-1} \right) \\ &=& 1 - \frac{n+1}{2^n}
\end{eqnarray}
$$
Confirming with Mathematica:
In[6]:= Probability[
un - u1 > 1/2, {u1, un} \[Distributed]
OrderDistribution[{UniformDistribution[], n}, {1,
n}]] // FullSimplify
Out[6]= 1 - 2^-n (1 + n)
In[7]:= Expectation[
un - u1, {u1, un} \[Distributed]
OrderDistribution[{UniformDistribution[], n}, {1,
n}]] // FullSimplify
Out[7]= (-1 + n)/(1 + n)
