Show $u_{xx}$ satisfies the PDE $u_t+cu_x=0$ 
Let $c$ be a constant and consider the following Cauchy problem associated with the inhomogeneous transport PDE $$u_t+cu_x=x, \ u(x,0)=x+2.$$
  a) Show $u_{xx}$ satisfies the homogeneous PDE and homogeneous initial data associated with the above Cauchy problem, i.e. $u_t+cu_x=0$ and $u(x,0)=0$.
b) Hence explain why $u_{xx}\equiv 0$
c) Hence, or otherwise, solve the original inhomogeneous Cauchy problem.

Honestly I don't really understand what the question is asking for. How can $u_{xx}$ satisfy a homogeneous, first-order PDE?
If we take 
\begin{align}
\frac{\partial}{\partial x}\left(u_t+cu_x\right)&=0 \\
u_{tx}+cu_{xx}&=0
\end{align}
 A: The notation in OP is somewhat confusing.
a) Let us follow the problem resolution as suggested by @littleO in the comments. Thus, we differentiate twice the PDE problem $u_t + c u_x = x$, $u(x,0) = x+2$ with respect to $x$:
$$
u_{txx} + c u_{xxx} = 0,
\qquad
u_{xx}(x,0) = 0 .
$$
Swapping derivatives yields $v_t + cv_x = 0$, $v(x,0) = 0$, where $v=u_{xx}$. Up to the notation for $u_{xx}$, this is exactly what is asked in a).
b) The solution of the Cauchy problem satisfied by $v$ can be deduced from the method of characteristics, and we have $v=u_{xx}\equiv 0$.
c) The method of characteristics is applied to $u_t + c u_x = x$, $u(x,0) = x+2$, which leads to
$$ u(x,t) = (x-ct) + 2 + (x-ct)t + \tfrac{1}{2} c t^2 . $$
Note that $u_{xx} \equiv 0$, as predicted in b).

Alternative approaches for c): 


*

*Set $w = u - \frac12 x^2/c$, which gives $w_t=u_t$ and $w_x=u_x - x/c$. Hence $w_t + cw_x =0$ with $w(x,0) = x+2 - \frac12 x^2/c$, which can be solved by applying the method of characteristics.

*Set $w = u - (x+2)$, which gives $w_t=u_t$ and $w_x=u_x - 1$. Hence $w_t + cw_x = c+x$ with $w(x,0) = 0$, which can be solved by applying the method of characteristics.
A: Hint.
From $u_t + c u_x = 0$
$$
\partial_x(u_t + c u_x) = u_{tx}+c u_{xx} = 0 \Rightarrow u_{xx} = -\frac 1c u_{tx}
$$
and now
$$
(u_{xx})_t + c(u_{xx})_x = \left(-\frac 1c u_{tx}\right)_t + c\left(-\frac 1c u_{tx}\right)_x = -\frac 1c\partial_t\partial_x\left(u_t+cu_x\right) = 0
$$
Also making $U(x,t) = u(x,t)+\frac{x^2}{2c}$ we have
$$
U_t+c U_x = 0\Rightarrow U(x,t) = \phi(x-c t)\Rightarrow u(x,t) = \phi(x-ct)-\frac{x^2}{2c}
$$
but $u(x,0) = x+2$ hence
$$
x+2 = \phi(x) - \frac{x^2}{2c}
$$
and finally
$$
u(x,t) = \frac{(x-c t)^2}{2c}+(x-c t)+2 -\frac{x^2}{2c}
$$
