Question on moment generating function
Part a. The MGF of $Y$ is, for any $b<1/2$, \begin{align} M_Y(b) &= E[\exp(-2b \ln(X))]\\ &=E[X^{-2b}]\\ &=\int_0^1 x^{-2b}dx\\ &=\frac{1}{1-2b}. \end{align}
Part b. The MGF of $Z$ satisfies $$M_Z(b)=E[\exp(b(Y_1+Y_2))]=E[\exp(bY_1)]E[\exp(bY_2)]=M_Y(b)^2=\left(\frac{1}{1-2b}\right)^2,$$ where I used that $Y_1$ and $Y_2$ are independent and distributed as $Y$. Thus, $Z$ has the MGF of a chi-square distribution with $r=4$ degrees of freedom. Since the MGF uniquely determines the distribution, $Z$ has density $$f_Z(z)=\frac{1}{4}x\exp(-x/2),\; x>0.$$