Consider that
$$P(X>Y)=E\left[\mathbf1_{X>Y}\right]$$
, where $\mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$E\left[g(X,Y)\right]=\iint g(x,y)f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
\begin{align}
P(X>Y)&=\iint \mathbf1_{x>y}\,e^{-x}\mathbf1_{x>0}\mathbf1_{1<y<2}\,\mathrm{d}x\,\mathrm{d}y
\\&=\iint \mathbf1_{x>y,\,x>0,1<y<2}\,e^{-x}\,\mathrm{d}x\,\mathrm{d}y
\end{align}
I answered quite the same question here. The only differences are that in the original post the distribution was $Exp(1)$ and the requeste distribution was the law of $Z_j=Y_{(j+1)}-Y_{(j)}$
Unfortunately this forum is an Italian forum...but I think you can understand the solution because formulas are formulas....in any language.
On the contrary please advice and I will repost the solution in English
Solution details
Just to simplify the notation, let's set $Y$ and $X$ the two order statistics and $Z=Y-X$ the rv we are looking for distribution ($Y>X$).
Using the known result on the joint density of two Order Statistics we get
$$f_{XY}(x,y)=\frac{n!}{(j-2)!(n-j)!}(1-e^{-\lambda x})^{j-2}e^{-\lambda(n-j)y}\lambda e^{-\lambda x}\lambda e^{-\lambda y}\cdot\mathbb{1}_{(x;\infty)}(y)$$
In order to get the CDF of the rv $Z=Y-X$ we have to integrate the joint density over the desired support, say:
$$F_Z(z)=\frac{n!}{(j-2)!(n-j)!}\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda e^{-\lambda x} \Bigg[ \int_x^{x+z}\lambda e^{-\lambda(n-j+1)y} dy\Bigg] dx$$
The integral is not difficult...it is enough to sove the inner integral in $dy$ (very easy) and get out of the sign of integral all what does not depend on x.
You get
$$\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda (e^{-\lambda x})^{n-j+2}dx$$
Now substitute
$1-e^{-\lambda x}=u$
and you get
$$F_Z(z)=\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\underbrace{\int_0^1 u^{j-2}(1-u)^{n-j+1}du}_{=Beta(j-1;n-j+2)}=1-e^{-\lambda(n-j+1)z}$$
As
$$Beta(j-1;n-j+2)=\frac{\Gamma(j-1)\Gamma(n-j+2)}{\Gamma(n+1)}=\frac{(j-2)!(n-j+1)!}{n!}$$
That is exactly the CDF they ask you to show.
Best Answer
I'm assuming $X,Y,Z$ are independent.
If $u \le v$ then $$P(U > u, V > v) = P(X > u, Y > v, Z > v) = P(X>u) P(Y>v) P(Z > v) = e^{-\lambda u - \mu v - \gamma v}.$$ The other case $u \ge v$ is similar.