-UPDATED SOLUTION from sasha's hint-
Thanks for checking my solution and see if i made any conceptual error.
the question http://dl.dropbox.com/u/5681270/Screen%20Shot%202011-11-09%20at%201.40.47%20PM.png
(i)
$f_{x}(x,y) = \int_{0}^{\infty} e^{-y} dy = 1 $ (uniform in 0 to 1)
$f_{y}(x,y) = \int_{0}^{1} e^{-y} dx = e^{-y} $ (exponential with unit mean) from the above, $f(x,y) = f_{X}(x)f_{Y}(y)$, therefore they are independent$F_v = P(min{X,Y}>v) = P(X>v)*P(Y>v) = (1-F_{x}(v))(1-F_{y}(v))$
$F_{X}(v) = v$ and $F_{Y}(v) = 1-e^{-v}$
therefore, $\frac{dF_{V}(v)}{dv}$
$ f_{min{X,Y}}(v) = \left\{ \begin{array}{lr} (v-2)e^{-v}-1 & : 0
(ii)
note that 2X has uniform distribution over $(o,2)$,
by convolution theorem,
$f_{2X+Y}(w) = \int_{-\infty}^{\infty}f_{2x}(x)*f_{Y}(w-x)dx$
$= \int_{0}^{2}\frac{1}{2}f_{Y}(w-x)dx$ since $f_{2x}(x) = \frac{1}{2}$ in $(0,2)$
as x goes from 0 to 2, y goes from w to w-2,
$= \frac{1}{2} \int_{w-2}^{w} f_{Y}(z)dz$
from here, we'll have 2 cases, (a) $0 < w < 2$ (b) $w > 2$
in (a)
$f_{2X + Y}(w) = \int_{0}^{w} e^(-z) dz = 1 - e^{-w}$
in (b)
$f_{2X + Y}(w) = \int_{w-2}^{w} e^(-z) dz = e^{-w - 2} - e^{-2}$
therefore, $ f_{2X + Y}(w) = \left\{ \begin{array}{lr} 1 - e^{-w} & : 0 < w < 2 \\ e^{-w - 2} - e^{-2} & : w > 2 \\ 0 & : otherwise \end{array} \right. $
Didier, did i get it right this time round?