I am able to do parts a) and b). I think that the other three parts are similiar.
For part c) I get that
$\displaystyle u(x,t)=\frac{1}{2}\int_0^{t} \int_{\infty}^{\infty}f(\xi,s)\Big[H(t-s-x+\xi)-H(-t+s-x+\xi) \Big] d\xi ds$
$\displaystyle =\frac{1}{2}\int_0^{t} \int_{\infty}^{\infty}\delta(\xi)\delta(s-1)\Big[H(t-s-x+\xi)-H(-t+s-x+\xi) \Big] d\xi ds$
$\displaystyle =\frac{1}{2}\int_0^{t} \delta(s-1)\Big[H(t-s-x)-H(-t+s-x) \Big] d\xi ds$
but I am not sure how this is evaluated. Surely if
$\begin{split}u(x,t)&=&\frac{1}{2}\int_0^{t}\delta(s-1)[H(t-s-x)-H(-t+s-x)]ds \\&=& \frac{1}{2}[H(t-1-x)-H(-t+1-x)]\end{split}$
then we need $1 \leq t$? Which I can see no indication of being true. I believe that understanding this issue will allow to solve the last part of the question. I cannot see how to change the limits in the last part.
Can someone please give a worked solution to parts c) and e). Offering a bounty in return.
c) Picking up where you left off, your issue lies with
$$u(x,t) = \frac{1}{2} \int_0^t \delta(s-1) \left [ H(t-s-x) - H ( -t + s -x) \right ] ds $$
If $t > 1$, then the delta function does it's thang and picks out $s =1$, since the integral covers that point. If $t \leq 1$ then the delta function acts as the null map. Thus we have:
$$ u(x,t) = \left \{ \begin{array}{cc} \frac{1}{2} \left ( H( t-1 - x) - H(-t +1-x) \right) & t>1 \\ 0 & t \leq 1 \end{array} \right. $$
Edit: Forgot about the graph! If we want to graph $u(x,2)$, we plug in to see:
$$u(x,2) = \frac{1}{2} \left( H ( 1 -x) - ( -1 -x) \right )= \frac{1}{2} H(1-|x|)$$
It's just a little box from $-1<x<1$ with height $1/2$
e)Use the formula you found with $f = p(x) \delta(t)$:
\begin{align*} u(x,t) =& \frac{1}{2} \int_0^t \int_{-\infty} ^ { \infty} p(\xi) \delta(s) \left [ H (t-s -x +\xi) - H( -t +s-x+ \xi) \right ] d\xi ds \\ =& \frac{1}{2} \int_{-\infty} ^ { \infty} p(\xi) \left [ H (t -x +\xi) - H( -t-x+ \xi) \right ] d\xi ds \end{align*}
Notice that if $t-x + \xi >0$ and $ - t -x + \xi>0$ the integral is zero. Thus we obtain $\xi > x+t$ and $\xi > x -t$ are our bounds for integration. In this range we have that:
$$ u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} p(\xi) d\xi $$
If $p(x) = H(x) e^x$, then we have
$$ u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} H(\xi) e^\xi d\xi = \left \{ \begin{array}{cc} \frac{1}{2} \int_{x-t}^{x+t} e^\xi d\xi & x > t \\ 0 & x<t \end{array} \right. =\left \{ \begin{array}{cc} e^x \sinh(t) & x > t \\ 0 & x<t \end{array} \right. $$
Hopefully that clears up the confusion!