Note that you are using the tail (complement CDF) $P(X > x)$ already.
The second of your starting equations
Now we can setup two equations with two unknowns. \begin{eqnarray*}
\Big( \frac{x_m}{100} \Big) ^ \alpha &=& 0.5 \\
\Big( \frac{x_m}{1000} \Big) ^ \alpha &=& \color{red}{0.9} \\
\end{eqnarray*}
should be just directly the "top $10\%$"
$$ \Big( \frac{x_m}{1000} \Big) ^ \alpha = 0.1$$
The solution is $x_m \approx 37.095689$ and $\alpha \approx 0.69897$. The left end cutoff $x_m$ is now a reasonably low value (hmm? totally a different universe from ours), and $\alpha > 0 $ gives the typical $1/x$ like shape.
The minimum of the top $2\%$ is exactly $10^4$K, and this is not a coincidence. By having the top $10\%$ to be ten times (in wealth) the top $50\%$, you are setting a power law that goes 10 times in wealth when you go one-fifth in top percentage. Thus $2\%$ is another 10 times of the $10\%$.
$(a)$ is not correct. The CDF is non-constant over $\Bbb R$ so the PDF is not $0$ for $x<0$. It is $e^{-x-e^{-x}}$ over all of $\Bbb R$.
$(b)$ is okay.
Your answer to $(c)$ is okay because you have made an error here which offsets the error you made in part $(a)$. Note that in $(a)$, you took $X\ge0$ so $Y=e^{-X}\le1$. Thus the "PDF" you should have got is $e^{-y},0<y\le1$, which is not a PDF at all since it doesn't integrate to give $1$. The error you made here is that you didn't find the range of $Y$ according to the range of $X$ you wrote in $(a)$ and automatically assumed $Y\ge0$.
With the correct bounds on $X$ i.e. $-\infty<X<\infty$, $Y$ would vary from $0\to\infty$, which corresponds to the exponential dsitribution with parameter $1$.
So the PDF is$$f(y)=\begin{cases}e^{-y},&y>0\\0,&y\le0\end{cases}$$The CDF is$$F(y)=\int_{-\infty}^yf(u)du=\begin{cases}0,&y\le0\\\int_\color{red}0^yf(u)du=1-e^{-y},&y>0\end{cases}$$
The error you have made in the CDF approach is that $Y\le y$ does not translate to $X\le g^{-1}(y)$ for $g$ which is not strictly increasing. In our case, $Y\le y\iff e^{-X}\le y\iff -X\le\ln y\iff X\color{red}\ge-\ln y=g^{-1}(y)$.
So $P(Y\le y)=\begin{cases}0,&y\le0\\P(X\ge-\ln y),&y>0\end{cases}\\=\begin{cases}0,&y\le0\\1-P(X\le-\ln y),&y>0\end{cases}\\=\begin{cases}0,&y\le0\\1-F_X(-\ln y),&y>0\end{cases}$
which gives the same answer.
Best Answer
If $k=0$ then $Y$ is identically $0$. If $k\gt 0$, then for suitable $y$ we have $$F_Y(y)=\Pr(Y\le y)=\Pr(kX\le y)=\Pr(X\le y/k)=1-\left(\frac{kx_m}{y}\right)^{\alpha}.$$ The suitable $y$ are where $\frac{y}{k}\ge x_m$. Elsewhere we have $F_Y(y)=0$.
The density function $f_Y(y)$ can now be obtained by differentiating.
Alternately, look up the formula for $f_Y(y)$ given by the method of transformations, and integrate to find $F_Y(y)$.