Your claimed result is not true, which probably explains why you're having trouble seeing it.
For simplicity I'll let $a = 0, b = 1$. Results for general $a$ and $b$ can be obtained by a linear transformation.
Let $X_1, \ldots, X_n$ be independent uniform $(0,1)$; let $Y$ be their minimum and let $X$ be their maximum. Then the probability that $X \in [x, x+\delta x]$ and $Y \in [y, y+\delta y]$, for some small $\delta x$ and $\delta y$, is
$$ n(n-1) (\delta x) (\delta y) (x-y)^{n-2} $$
since we have to choose which of $X_1, \ldots, X_n$ is the smallest and which is the largest; then we need the minimum and maximum to fall in the correct intervals; then finally we need everything else to fall in the interval of size $x-y$ in between. The joint density is therefore $f_{X,Y}(x,y) = n(n-1) (x-y)^{n-2}$.
Then the density of $Y$ can be obtained by integrating. Alternatively, $P(Y \ge y) = (1-y)^n$ and so $f_Y(y) = n(1-y)^{n-1}$.
The conditional density you seek is then
$$ f_{X|Y}(x|y) = {n(n-1) (x-y)^{n-2} \over n(1-y)^{n-1}} == {(n-1) (x-y)^{n-2} \over (1-y)^{n-1}}. $$
where of course we restrict to $x > y$.
For a numerical example, let $n = 5, y = 2/3$. Then we get $f_{X|Y}(x/y) = 4 (x-2/3)^3 / (1/3)^4 = 324 (x-2/3)^3$ on $2/3 \le x \le 1$. This is larger near $1$ than near $2/3$, which makes sense -- it's hard to squeeze a lot of points in a small interval!
The result you quote holds only when $n = 2$ -- if I have two IID uniform(0,1) random variables, then conditional on a choice of the minimum, the maximum is uniform on the interval between the minimum and 1. This is because we don't have to worry about fitting points between the minimum and the maximum, because there are $n - 2 = 0$ of them.
This is really the same as drawing the pair $(x,y)$ uniformly from inside the annulus bounded by concentric circles of radii $r_\min$ and $r_\max$. If $s$ is between $r_\min$ and $r_\max$ then the probability that $r\le s$ is the area between the circle of radius $s$ and the circle of radius $r_\min$ divided by the whole area of the annulus. I.e.
$$
\Pr(r \le s) = \frac{\text{area of annulus between radii }r_\min\text{ and }s}{\text{area of annulus between radii }r_\min\text{ and }r_\max} = \frac{\pi s^2 - \pi r_\min^2}{\pi r_\max^2 - \pi r_\min^2} = \frac{s^2-r_\min^2}{r_\max^2-r_\min^2}.
$$
So that's the CDF as a function of $s$. Differentiate it to get the PDF, and the numerator becomes $2s$ while the denominator stays as it was. Call the CDF (capital) $F$ and density (lower-case) $f$, and then the expected value is
$$
\alpha\int_{r_\min}^{r_\max} \frac{1}{s^2} f(s)\;ds.
$$
Best Answer
Your calculations are correct! Just for the sake of completeness, you have to write
$$F_M(t)=t^2\mathbb{1}_{(0;1)}(t)+\mathbb{1}_{[1;\infty)}(t)$$
...and similar notation for the other CDF's. This because the CDF is defined over all $\mathbb{R}$
Why the law of L and T are the same? As you did to find $F_T$ do the same drawing to find $F_L$ and you will realize that the integration area is the same