Use the method of transformations for both exercises.
a.
[tex]\begin{cases}U=5X\\V=\frac{4X}Y\end{cases}\implies\begin{cases}X=\frac U5\\Y=\frac{4U}{5V}\end{cases}[/tex]
The Jacobian for this transformation is
[tex]J=\begin{bmatrix}X_U&X_V\\Y_U&Y_V\end{bmatrix}=\begin{bmatrix}\frac15&0\\\frac4{5V}&-\frac{4U}{5V^2}\end{bmatrix}\implies|\det J|=\frac{4U}{25V^2}[/tex]
Then the joint density of [tex]U[/tex] and [tex]V[/tex] is
[tex]f_{U,V}(u,v)=f_{X,Y}\left(\dfrac U5,\dfrac{4U}{5V}\right)|\det J|[/tex]
Both [tex]X[/tex] and [tex]Y[/tex] are i.i.d with density
[tex]f_R(r)=\begin{cases}1&\text{for }0<r<1\\0&\text{otherwise}\end{cases}[/tex]
(where [tex]R[/tex] is either [tex]X[/tex] or [tex]Y[/tex]) so we have
[tex]f_{U,V}(u,v)=\begin{cases}\frac{4u}{25v^2}&\text{for }0<\frac u5<1\text{ and }0<\frac{4u}{5v}<1\\0&\text{otherwise}\end{cases}[/tex]
We can clean up the support a bit:
[tex]0<\dfrac{4u}{5v}<1\implies\dfrac{5v}{4u}<0\text{ or }\dfrac{5v}{4u}>1[/tex]
The first case suggests either [tex]U[/tex] or [tex]V[/tex] can be negative, but with [tex]U=5X[/tex], that is impossible. This leaves us with
[tex]\dfrac{5v}{4u}>1\implies v>\dfrac{4u}5[/tex]
so the joint PDF is
[tex]f_{U,V}(u,v)=\begin{cases}\frac{4u}{25v^2}&\text{for }0<\frac u5<1\text{ and }v>\frac{4u}5\\0&\text{otherwise}\end{cases}[/tex]
b.
[tex]\begin{cases}U=3X+Y\\V=\frac{5X}{X+Y}\end{cases}\implies\begin{cases}X=\frac{UV}{5+2V}\\Y=\frac{(5-V)U}{5+2V}\end{cases}[/tex]
The Jacobian is
[tex]J=\begin{bmatrix}\frac V{5+2V}&\frac{5U}{(5+2V)^2}\\\frac{5-V}{5+2V}&-\frac{15UV}{(5+2V)^2}\end{bmatrix}\implies|\det J|=\dfrac{5U}{(5+2V)^2}[/tex]
and so the joint PDF is
[tex]f_{U,V}(u,v)=f_{X,Y}\left(\frac{uv}{5+2v},\frac{(5-v)u}{5+2v}\right)|\det J|[/tex]
[tex]f_{U,V}(u,v)=\begin{cases}\frac{5u}{(5+2v)^2}&\text{for }0<\frac{uv}{5+2v}<1,0<\frac{(5-v)u}{5+2v}<1\\0&\text{otherwise}\end{cases}[/tex]
We have
[tex]0<\dfrac{uv}{5+2v}<1\implies0<uv<5+2v[/tex]
and
[tex]0<\dfrac{(5-v)u}{5+2v}<1\implies0<5u-uv<5+2v[/tex]
which together tell us that
[tex]0<5u<2(5+2v)\implies0<u<2+\dfrac{4v}5[/tex]
As for [tex]v[/tex], we know that at most we can have [tex]X=Y=1[/tex], so that [tex]V[/tex] can have a maximum value of 5/2, so we can simplify the support to write the PDF as
[tex]f_{U,V}(u,v)=\begin{cases}\frac{5u}{(5+2v)^2}&\text{for }0<u<2+\frac{4v}5,0<v<\frac52\\0&\text{otherwise}\end{cases}[/tex]
