Probability that $max(X_1, ldots, X_n) - min(X_1, ldots, X_n) leq 0.5$

Let $Z_n$ equal the maximum, and $Z_1$ the minimum of ${{X_k}}_{kin{1..n}}$, which is a sample of $n$ iid random variables with probability density function $f_{small X}(x)$ and cummulative distribution function $F_{small X}(x)$.

$$begin{align}f_{small X}(x)&=mathbf 1_{xin(0..1)}\[2ex]F_{small X}(x)&=xmathbf 1_{xin(0..1)}+mathbf 1_{xin[1..infty)}end{align}$$

Then we find

$$begin{align}f_{small Z_1,Z_n}(s,t) &= dfrac{n!}{,1!,(n-2)!,1!,} f_{small X}(s)bigl(F_{small X}(t)-F_{small X}(s)bigr)^{n-2}f_{small X}(t)mathbf 1_{0leq sleq tleq 1}\[1ex]&=n(n-1)(t-s)^{n-1}mathbf 1_{0leq sleq tleq 1}\[2ex]f_{small (Z_n-Z_1)}(z)&=int_0^{1-z} f_{small Z_1,Z_2}(s,s+z)~mathrm ds\[2ex]F_{small (Z_n-Z_1)}(z)&=int_0^z f_{small (Z_n-Z_1)}(u)~mathrm d u\[1ex]&= n(n-1)int_0^{z}int_0^{1-u} u^{n-1}~mathrm d s~mathrm d uend{align}$$

Let $A = 1-min(X_1,ldots,X_n)$ and $B = max(X_1,ldots,X_n)$. The joint CDF of $A,B$ is given by:

begin{align*} F(a,b) &= P{A le a, B le b} \ &= P{min(X_1,ldots,X_n) ge 1-a text{and} max(X_1,ldots,X_n) le b} \ &= P{1-a le X_k le b text{for} k = 1,ldots,n} \ &= prod_{k = 1}^{n}P{1-a le X_k le b} \ &= prod_{k = 1}^{n}(b-(1-a))I(b ge 1-a) \ &= (b-(1-a))^nI(b ge 1-a) end{align*}

for $0 le a,b le 1$.

The joint PDF of $A,B$ can be found by computing $f(a,b) = dfrac{partial^2F}{partial a partial b}(a,b)$, and then you can compute $P{max(X_1,ldots,X_n)-min(X_1,ldots,X_n) le tfrac{1}{2}} = P{B-(1-A) le tfrac{1}{2}}$ by integrating $f(a,b)$ over the appropriate subset of $[0,1] times [0,1]$.

I think you're coming at this from a slightly wrong direction; importantly, $Z_1$ and $Z_n$ aren't independent so knowing their individual distributions doesn't help you. For instance, in the $n=2$ case, $Z_1geq frac12$ with probability $frac14$ and with the same probability $Z_2leq frac12$, but these two events can never happen simultaneously.

Instead, suppose that $Z_1=z$. Then all of the other $X_i$ are equidistributed in $[z, 1]$ (why?). So your probability for this value of $Z_1$ is simply $displaystyleprod_{i, X_ineq Z_1} Pleft(X_i leq min(1, z+0.5) | zleq X_iright)$. And since the $X_i$ are independent, this is just $displaystyleleft(Pleft(X_i leq min(1, z+0.5) | zleq X_iright)right)^{n-1}$. Then you'll have to integrate this over the distribution of $Z_1$ (which you should be able to find with a sort of symmetry argument.)