DPLL algorithm:

\scalebox{1}{
\begin{tabular}{|l|c|c|c|c|c|c|c|c|}
\hline
Step                                & 1         & 2         & 3           & 4             & 5                    & 6          & 7         & 8               \\ \hline \hline
Decision Level                      & 0         & 0         & 0           & 1             & 1                    & 0          & 0         & 0               \\ \hline
Assignment                          & -         & $\lnot a$ & $\lnot a,c$ & $\lnot a,b,c$ & $\lnot a,b,c,e$      & -          & $\lnot b$ & $\lnot b, f$    \\ \hline
Cl. 1: $\lnot a,d$                  & 1         & \cmark    & \cmark      & \cmark        & \cmark               & 1          & 1         & 1               \\ \hline
Cl. 2: $\lnot d,c$                  & 2         & 2         & \cmark      & \cmark        & \cmark               & 2          & 2         & 2               \\ \hline
Cl. 3: $\lnot b,e$                  & 3         & 3         & 3           & $e$           & \cmark               & 3          & \cmark    & \cmark          \\ \hline
Cl. 4: $\lnot b,\lnot e$            & 4         & 4         & 4           & $\lnot e$     & $\{\}$ \xmark        & 4          & \cmark    & \cmark          \\ \hline
Cl. 5: $b,f$                        & 5         & 5         & 5           & \cmark        & \cmark               & 5          & $f$       & \cmark          \\ \hline
Cl. 6: $b,\lnot f$                  & 6         & 6         & 6           & \cmark        & \cmark               & 6          & $\lnot f$ & $\{\}$ \xmark   \\ \hline
\textcolor{gray}{Cl. 7: $\lnot b$}  & -         & -         & -           & -             & learned $\lnot b$    & 7          & \cmark    & \cmark          \\ \hline \hline
BCP                                 & -         & -         & -           & $e$           & -                    & $\lnot b$  & $f$       & -               \\ \hline
PL                                  & $\lnot a$ & $c$       & -           & -             & -                    & -          & -         & -               \\ \hline
Decision                            & -         & -         & $b$         & -             & -                    & -          & -         & UNSAT           \\ \hline
\end{tabular}}

Ad 5:

\begin{center}
\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
    thick,base node/.style={circle,draw,minimum size=20pt}, real node/.style={double,circle,draw,minimum size=20pt}]

    \node[base node] (1) {$b$};
    \node[base node] (2) [above right of=1] {$e$};
    \node[base node] (3) [below right of=1] {$\lnot e$};
    \node[base node] (4) [above right of=3] {$\bot$};
    \path[]
        (1) edge [] node {$3$} (2)
            edge [] node {$4$} (3)
        (2) edge [] node {} (4)
        (3) edge [] node {} (4);
\end{tikzpicture}
\end{center}

\begin{prooftree}
\AxiomC{$3. \; \lnot b \lor e$}
\AxiomC{$4. \; \lnot b \lor \lnot e$}
	\BinaryInfC{$\lnot b$}
\end{prooftree}

Ad 8:

\begin{center}
\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
    thick,base node/.style={circle,draw,minimum size=20pt}, real node/.style={double,circle,draw,minimum size=20pt}]

    \node (0) {};
    \node[base node] (1) [right of=0] {$\lnot b$};
    \node[base node] (2) [above right of=1] {$f$};
    \node[base node] (3) [below right of=1] {$\lnot f$};
    \node[base node] (4) [above right of=3] {$\bot$};
    \path[]
        (0) edge [] node {$7$} (1)
        (1) edge [] node {$5$} (2)
            edge [] node {$6$} (3)
        (2) edge [] node {} (4)
        (3) edge [] node {} (4);
\end{tikzpicture}
\end{center}

\begin{prooftree}
\AxiomC{$5. \; b \lor f$}
\AxiomC{$6. \; b \lor \lnot f$}
	\BinaryInfC{$b$}
	\AxiomC{$7. \; \lnot b$}
		\BinaryInfC{$\bot$}
\end{prooftree}