Looks like there are two distinct problems here.
[tex]P(\ln H\ge z)=P(\ln((X+2)Y)\ge z)=P((X+2)Y\ge e^z)=P\left(Y\ge\dfrac{e^z}{X+2}\right)[/tex]
Conditioning on [tex]X[/tex] involves interpreting this probability for all possible fixed values of [tex]X=x[/tex]:
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\text{ and }X=x\right)[/tex]
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_xP\left(Y\ge\dfrac{e^z}{X+2}\mid X=x\right)P(X=x)[/tex]
We have [tex]x\in(0,1)[/tex], and over this domain [tex]P(X=x)=1[/tex]. Since [tex]X=x[/tex] is fixed, we can omit the conditioning notation to leave us with
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1P\left(Y\ge\dfrac{e^z}{x+2}\right)\,\mathrm dx[/tex]
[tex]Y[/tex] has CDF
[tex]P(Y\le y)=F_Y(y)=\begin{cases}0&\text{for }y<0\\y&\text{for }y\in(0,1)\\1&\text{for }y>1\end{cases}[/tex]
Then we can write our integral as
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1\left(1-F_Y\left(\frac{e^z}{x+2}\right)\right)\,\mathrm dx[/tex]
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\displaystyle\int_0^1\left(1-\frac{e^z}{x+2}\right)\,\mathrm dx[/tex]
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=x-e^z\ln(x+2)\bigg|_{x=0}^{x=1}[/tex]
[tex]P\left(Y\ge\dfrac{e^z}{X+2}\right)=\boxed{1-e^z(\ln2-\ln3)}[/tex]
Using the method of distribution functions, we have
[tex]F_Z(z)=P(Z\le z)=P(F_X(X)\le z)=P(X\le {F_X}^{-1}(z))=F_X({F_X}^{-1}(z))=z[/tex]
[tex]\implies\boxed{f_Z(z)=\begin{cases}1&\text{for }z\in(0,1)\\0&\text{otherwise}\end{cases}}[/tex]
