DPLL algorithm:

\scalebox{1}{
\setlength\tabcolsep{3pt}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|}
\hline
Step                        & 1        & 2        & 3           & 4             & 5                    & 6   & 7         & 8          & 9                 & 10                          \\ \hline \hline
Decision Level              & 0        & 1        & 1           & 1             & 1                    & 0   & 0         & 0          & 0                 & 0                           \\ \hline
Assignment                  & -        & $\lnot a$& $\lnot a,c$ & $\lnot a,b,c$ & $\lnot a,b,c,\lnot d$& -   & $a$       &$a,\lnot b$ &$a,\lnot b,\lnot c$& $a,\lnot b,\lnot c,\lnot e$ \\ \hline
Cl. 1: $\lnot a,\lnot b$    & 1        & \cmark   & \cmark      & \cmark        & \cmark               & 1   & $\lnot b$ & \cmark     & \cmark            & \cmark                      \\ \hline
Cl. 2: $ a,c$               & 2        & $c$      & \cmark      & \cmark        & \cmark               & 2   & \cmark    & \cmark     & \cmark            & \cmark                      \\ \hline
Cl. 3: $b,\lnot c$          & 3        & 3        & $b$         & \cmark        & \cmark               & 3   & 3         & $\lnot c$  & \cmark            & \cmark                      \\ \hline
Cl. 4: $\lnot b,d$          & 4        & 4        & 4           & $d$           & $\{\}$ \xmark        & 4   & 4         & \cmark     & \cmark            & \cmark                      \\ \hline
Cl. 5: $\lnot c,\lnot d$    & 5        & 5        & $\lnot d$   & $\lnot d$     & \cmark               & 5   & 5         & 5          & \cmark            & \cmark                      \\ \hline
Cl. 6: $c,e$                & 6        & 6        & \cmark      & \cmark        & \cmark               & 6   & 6         & 6          & $e$               & $\{\}$ \xmark               \\ \hline
Cl. 7: $c,\lnot e$          & 7        & 7        & \cmark      & \cmark        & \cmark               & 7   & 7         & 7          & $\lnot e$         & \cmark                      \\ \hline
\textcolor{gray}{Cl. 8: $a$}& -        & -        & -           & -             & learned $a$          & 8   & \cmark    & \cmark     & \cmark            & \cmark                      \\ \hline \hline
BCP                         & -        & $c$      & $b$         & $\lnot d$     & -                    & $a$ & $\lnot b$ & $\lnot c$  & $\lnot e$         & -                           \\ \hline
PL                          & -        & -        & -           &  -            & -                    & -   &     -     & -          & -                 & -                           \\ \hline
Decision                    & $\lnot a$&   -      &     -       &       -       &    -                 & -   & -         &    -       &   -               & 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) {$\lnot a$};
    \node[base node] (2) [right of=1]{$c$};
    \node[base node] (3) [below right of=2] {$\lnot d$};
    \node[base node] (4) [above right of=2] {$b$};
    \node[base node] (5) [right of=4] {$d$};
    \node[base node] (6) [below right of=5] {$\bot$};
    \path[]
        (1) edge [] node {$2$} (2)
        (2) edge [] node {$5$} (3)
            edge [] node {$3$} (4)
        (4) edge [] node {$4$} (5)
        (5) edge [] node {} (6)
        (3) edge [] node {} (6);
\end{tikzpicture}
\end{center}

\begin{prooftree}
\AxiomC{$4. \; \lnot b \lor d$}
\AxiomC{$5. \; \lnot c \lor \lnot d$}
	\BinaryInfC{$\lnot b \lor \lnot c$}
	\AxiomC{$3. \; b \lor \lnot c$}
		\BinaryInfC{$\lnot c$}
		\AxiomC{$2. \; a \lor c$}
			\BinaryInfC{$a$}
\end{prooftree}

Ad 10:

\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]{$a$};
    \node[base node] (2) [right of=1]{$\lnot b$};
    \node[base node] (3) [right of=2] {$\lnot c$};
    \node[base node] (4) [above right of=3] {$\lnot e$};
    \node[base node] (5) [below right of=3] {$e$};
    \node[base node] (6) [above right of=5] {$\bot$};
    \path[]
        (0) edge [] node {$8$} (1)
        (1) edge [] node {$1$} (2)
        (2) edge [] node {$3$} (3)
        (3) edge [] node {$7$} (4)
            edge [] node {$6$} (5)
        (4) edge [] node {} (6)
        (5) edge [] node {} (6);
\end{tikzpicture}
\end{center}

\begin{prooftree}
\AxiomC{$6. \; c \lor e$}
\AxiomC{$7. \; c \lor \lnot e$}
	\BinaryInfC{$c$}
	\AxiomC{$3. \; b \lor \lnot c$}
		\BinaryInfC{$b$}
		\AxiomC{$1. \; \lnot a \lor \lnot b$}
			\BinaryInfC{$\lnot a$}
			\AxiomC{$8. \; a$}
			    \BinaryInfC{$\bot$}
\end{prooftree}