$$\mathbb{P}(X-Y<t\mid X>Y) =\frac{\mathbb{P}(X-Y<t,X>Y)}{\mathbb{P}(X>Y)} =\mathbb{P}(X-Y<t)$$ since $X(\omega)>Y(\omega),\forall\omega\in\Omega

$$\mathbb{P}(X-Y<t)=\int_{0}^{\infty}dy\int_{0}^{y+t}f_{(X,Y)}(x,y)dx$$ where $f_{(X,Y)}(x,y)=p_{1}f_{Exp(\alpha_{1})}\ast\cdots\ast p_{n}f_{Exp(\alpha_{n})}\ast q_{1}f_{Exp(\beta_{1})}\ast\cdots\ast q_{m}f_{Exp(\beta_{m})}$  (which follows (2.18) and (2.19) in Akkouchi's paper)with $\sum_{i}p_{i}=\sum_{j}q_{j}=1$.

and we know that $f_{Exp(\alpha)}(x)=\alpha e^{-\alpha x}$. The nontrivial part of this problem is to derive a closed form of the integral, which I doubt it is possible for arbitrary $n,m$ since such a hyper geometric function is not always simplifiable.

$$\mathbb{P}(X-Y<t\mid X>Y) =\frac{\mathbb{P}(X-Y<t,X>Y)}{\mathbb{P}(X>Y)} =\mathbb{P}(X-Y<t)$$ since $X(\omega)>Y(\omega),\ \forall\omega\in\Omega $.

$$\mathbb{P}(X-Y<t)=\int_{0}^{\infty}dy\int_{0}^{y+t}f_{(X,Y)}(x,y)dx$$ where $f_{(X,Y)}(x,y)=p_{1}f_{Exp(\alpha_{1})}\ast\cdots\ast p_{n}f_{Exp(\alpha_{n})}\ast q_{1}f_{Exp(\beta_{1})}\ast\cdots\ast q_{m}f_{Exp(\beta_{m})}$ 

(which follows (2.18) and (2.19) in Akkouchi's paper) with $\sum_{i}p_{i}=\sum_{j}q_{j}=1$.

and we know that $f_{Exp(\alpha)}(x)=\alpha e^{-\alpha x}$. The nontrivial part of this problem is to derive a closed form of the integral, which I doubt it is possible for arbitrary $n,m$ since such a hyper geometric function is not always simplifiable.