(a), (c), (d) solved correctly. On attached picture (built with online plotter
http://math.rice.edu/~dfield/dfpp.html) where notations are a bit different ($(x,y)$ instead of $(t,x)$) one can see characteristics and initial line $t=0$ -- bold). The characteristics cross it no more than once (good, no contradiction) and initial data define solution on the characteristic which cross initial line (so $|x|>|t|$ which consists of two sectors)
(b), (e) contain glitches:
(b) AR believes that solution has the form $\phi \bigl( \frac{x^2+1}{t^2+1}\bigr)$ which would mean that the solution must be an even function with respect to $t$ and with respect to $x$ which is not necessary the case. In reality $\frac{x^2+1}{t^2+1}$ marks not the characteristic but the whole curve consisting of two disjoint components and on these components $u$ by no means is the same. Therefore absolutely correct answer is
In $k$-th sector ($\{x>|t|\}$, $\{x<-|t|\}$ as $k=1,3$ and $\{t>|x|\}$, $\{t<-|x|\}$ as $k=2,4$) general solution is
$u(x,t)=\phi _k \bigl( \frac{x^2+1}{t^2+1}\bigr)$ with $\phi_{1,3}(\xi)$ defined for $\xi>1$ and $\phi_{2,4}(\xi)$ defined for $0<\xi<1$.
(one can give other equivalent descriptions).
So, again, initial data define solutions in sectors $1,3$.
In (d) initial function$x^2$ is even with respect to $x$ and therefore the above remark has no effect and the answer is $\phi (\xi)=\xi -1$ and
\begin{equation*}
u(x,t)= \frac{x^2+1}{t^2+1}-1 =\frac{x^2-t^2}{t^2+1}.
\end{equation*}
In (e) initial function$x$ is odd with respect to $x$ and therefore the above remark works in full: $\phi_{1,3}(\xi)= \pm \sqrt{\xi-1}$ and
\begin{equation*}
u(x,t)= \frac{x^2+1}{t^2+1}-1 =\left\{\begin{aligned}
&\sqrt{\frac{x^2-t^2}{t^2+1}}\qquad &&x>|t|,\\
-&\sqrt{\frac{x^2-t^2}{t^2+1}}\qquad &&x<-|t|
\end{aligned}\right.
\end{equation*}
PS Quo usque tandem abutere, Catilina, patientia nostra?
To what length will you abuse our patience, Catiline?
Well, I meant, I am not going to look at your colourful scans anymore
