Need a worked solution to part d) Greens function wont fall out 
For part a) I get $\frac{d}{dx}\Big[ \frac{1}{x}\frac{dy}{dx}\Big] - \frac{3g}{x^3}=\frac{f(x)}{x}$
for part b) the two solutions are $1/x$ and $x^3$.
part c) is not hard to prove and I will not do so here
for part d) I get stuck.
I get that $Lg=\delta(x-\xi)$ and $g(1,\xi)=g(2,\xi)=0$
$\implies  g(x,\xi) =
\left\{
 \begin{array}{ll}
  \frac{A}{x}+Bx^3  & \mbox{if } 1<x<\xi \\
  \frac{c}{x}+Dx^3 & \mbox{if } \xi<x<2
 \end{array}
\right.$
plugging in the BC's I get
$\implies  g(x,\xi) =
\left\{
 \begin{array}{ll}
  \frac{A}{x}-Ax^3  & \mbox{if } 1<x<\xi \\
  \frac{C}{x}-\frac{Cx^3}{16} & \mbox{if } \xi<x<2
 \end{array}
\right.$
then using the continuity condition I get
$\frac{A}{\xi}-A{\xi}^3=\frac{C}{\xi}-\frac{C\xi^3}{16}$ which gives that
$A=\frac{C(16-\xi^4)}{16(1-\xi^4)}$ 
then using $\frac{1}{p(\xi)}\Big[g'(\xi^+,\xi)-g'(\xi^-,\xi)\Big]=1$ I get that
$\Big[ 3C\xi^2+\frac{16C}{\xi^2}\Big] - \Big[C \Big(\frac{\xi^4-16}{\xi^4-1}    \Big)\Big(3\xi^2+\frac{1}{\xi^2}\Big)\Big]=1$
$\implies 3C\xi^2+\frac{16C}{\xi^2} - 3C\xi^2 \Big(\frac{\xi^4-16}{\xi^4-1}    \Big)-C\Big(\frac{\xi^4-16}{\xi^4-1}\Big)\frac{1}{\xi^2}=1$
I would like a  worked solution to part d) in this question and in return I am offering a bounty.

The (a) and (b) are trivial to show. Let us start from (c). To find the Green's function, as you mentioned, we need to solve $Lg=\delta(x-\xi)$. For the sake of simplicity, I use the common form of the Green's function instead of solving a BVP and you can find relative theorems and procedures in some references (such as Wolfgang Walter, Ordinary Differential Equations P252 or the link given by @Jeb). The Green's function is given by:
$$
g(x,\xi)=\left\{
\begin{array}{l}
\dfrac{x^4\xi^4-16x^4-\xi^4+16}{60x\xi^2}\qquad x\leq \xi\\[3mm]
\dfrac{x^4\xi^4-16\xi^4-x^4+16}{60x\xi^2}\qquad x> \xi\\
\end{array}
\right.
$$
Then (c) is solved.
From a simple computation, namely the convolution of Green's function and RHS function $f(x)$, we have:
$$
y(x)=\int_1^2 g(x,\xi)f(\xi) d\xi=\int_1^x \dfrac{x^4\xi^4-16\xi^4-x^4+16}{60x\xi^3} d\xi+\int_x^2 \dfrac{x^4\xi^4-16x^4-\xi^4+16}{60x\xi^3} d\xi\\
=\dfrac{4x^4-15x^2-4}{60x}-\dfrac{x^4-16}{60x}=\dfrac{x^4-5x^2+4}{20x}
$$
Thus (d) is valid.
For (e), we should note that the $x^3$ and $1/x$ are still the independent solutions to the homogeneous problem $Ly=0$ in spite of the initial data at $x=1$ shift from $0$ to $c$. What is more, it easy to check that the $y(x)=-x/4$ is a particular solution to $f(x)=1/x$ in RHS (Hint: by plug $y(x)=kx$ in the ODE to solve a $k$). So the solution of $Ly=1/x$ with $y(1)=c$ and $y(2)=0$ can be written as
$$
y(x)=k_1x^3+\dfrac{k_2}{x}-\dfrac{x}{4}
$$
where $k_1$ and $k_2$ are constant to be determined. Using BC, we have:
$$
y(x) = \dfrac{1}{5x}+\dfrac{16c}{15x}+x^3(-\dfrac{c}{15}+\dfrac{1}{20})-\dfrac{x}{4}
$$
Solving $y'(x)|_{x=2}=0$ obtain $c=\dfrac{9}{32}$.
