Re 1., assume for the sake of simplicity that the $n$ points are i.i.d. in $(0,1)$ then their minimum $m$ and their maximum $M$ are such that, for every $x\leqslant y$ in$(0,1)$, $\mathbb P(x\leqslant m, M\leqslant y)=(y-x)^n$. Differentiating twice yields the density of $(m,M)$ as
$$f_{m,M}(x,y)=n(n-1)(y-x)^{n-2}\mathbf 1_{0\leqslant x\leqslant y\leqslant1}.
$$
Let $L=M-m$, then $L=z$ if $m=x$ and $M=x+z$ for some $x$ in $(0,1-z)$ hence the density of $L$ is
$$
f_L(z)=n(n-1)z^{n-2}(1-z)\mathbf 1_{0\leqslant z\leqslant1}.
$$
In terms of $\ell=x_{\mathrm{max}}-x_{\mathrm{min}}$ which corresponds to points in $(-a,a)$,
$$
f_\ell(t)=n(n-1)(2a)^{-n}t^{n-2}(2a-t)\mathbf 1_{0\leqslant t\leqslant 2a}.
$$
I do not know what 2. means.
Edit: When $n\to\infty$, the asymptotics of $\ell$ is as follows. Define
$$
R_n=n\left(1-\frac{\ell}{2a}\right),
$$
then $R_n$ converges in distribution to the sum of two independent standard exponential random variables (the gamma$(2,1)$ distribution), that is,
$$
f_{R_n}(r)\to r\mathrm e^{-r}\mathbf 1_{r\geqslant0}.
$$
Edit bis: The PDF $f_L$ yields the CDF of $L$ since, for every $0\leqslant z\leqslant1$,
$$
\mathbb P(L\leqslant z)=\int_0^zf_L(t)\mathrm dt=[nt^{n-1}-(n-1)t^n]_0^z=(n-(n-1)z)z^{n-1}.
$$
In particular, in the setting of the Edit to the question, for every $\delta\leqslant\frac12$ and every $n\geqslant2$,
$$
\mathbb P(x_{\mathrm{max}}-x_{\mathrm{min}}\leqslant\delta)=\mathbb P(L\leqslant\delta/(1-\delta))=(n-(2n-1)\delta)\delta^{n-1}/(1-\delta)^n.
$$
Edit ter: The confusion underlying part 2. might be the following. Let $(x_k)_{1\leqslant k\leqslant n}$ denote a i.i.d. sample uniform on $(-a,a)$, then the central limit theorem asserts that $\frac1{\sqrt{n}}\sum\limits_{k=1}^nx_k$ converges in distribution to a normal random variable, centered with variance $\mathbb E(x_1^2)=\frac13a^2$. For example, for every $b\geqslant0$,
$$
\mathbb P(-b\sqrt{n}\leqslant x_1+\cdots+x_n\leqslant b\sqrt{n})\to\sqrt{\frac2\pi}\int_0^{b\sqrt3/a}\mathrm e^{-x^2/2}\mathrm dx.
$$
But this concerns the empirical mean $\frac1n\sum\limits_{k=1}^nx_k$ of the sample, not its range $\ell=x_{\mathrm{max}}-x_{\mathrm{min}}$. The fluctuations of the range, as explained above, are described by an exponential distribution and not by a normal distribution.
