a.
We start from an arbitrary vertex $A_1$, and find one of its neighbour $A_2$. Since $\deg A_2 = 2$, $A_2$ has another neighbour $A_3$. Repeating the process, we must eventually find a vertex $A_k$ whose another neighbour is $A_1$ since there are finitely many vertices and all of $A_2, A_3, \dots, A_{k-1}$ already have 2 neighbours. Now, all vertices already have degree 2. Therefore, this must be the whole graph or otherwise $G$ cannot be connected. Thus, $A_1A_2\cdots A_k$ is a polygon.

b.
It suffices to show that $G$ has a $(\Delta_G + 1)$-colouring. Indeed, we can just assign an arbitrary colour to the vertices one by one such that each vertex has a different colour as all its neighbours which are already coloured. This is possible since there are $\Delta_G + 1$ colours but each vertex has degree at most $\Delta_G$.

c.
$G$ can be a cycle graph with an odd number of vertices and any complete graph.

Firstly, a cycle graph with an odd number of vertices satisfies $\chi_G = 3$ and $\Delta_G = 2$. Indeed, $\Delta_G = 2$ holds by definition. If at most 2 colours are used, there is a unique way to colour the vertices $A_1, A_2, \dots$ of the polygon $A_1A_2\cdots A_k$ such that $A_{j+1}$ has a different colour as $A_j$. But since $k$ is odd, $A_k$ has the same colour as $A_1$, contradiction. Therefore, at least 3 colours are needed. Since $\chi_G \le \Delta_G + 1 = 3$, we must have $\chi_G = 3 = \Delta_G + 1$.

Secondly, a complete graph with $k$ vertices satisfies $\chi_G = k$ and $\Delta_G = k - 1$.

Both are obvious since any pair of vertices is joined by an edge (hence all vertices have different colours).

d.
(Brook's theorem) We prove the result by induction on $\Delta_G$, and for each $\Delta_G$, we induct on the number $k$ of vertices.
The base case $\Delta_G = 2$ is trivial since the graph is just a path or a cycle. Suppose $\Delta_G \ge 3$. Now, the base case is $k = \Delta_G + 1$. As $G$ is not a complete graph, there is a vertex $A$ which has degree at most $\Delta_G - 1$. We can colour all other vertices in different colours ($\Delta_G$ colours are used). Then since $\deg A \le \Delta_G - 1$, we can always assign a colour to $A$ which is different from the colours of all its neighbours.

Now, consider the inductive step, where $k \ge \Delta_G + 2$.

Case 1. There exists a vertex $A$ such that $G - \{A\}$ is disconnected.
Suppose the connected components of $G - \{A\}$ are $G_1, G_2, \dots, G_m$. For each $G_j$, consider the connected subgraph $G'_j$ induced by $G$ containing all the vertices in $G_j$ together with $A$. Since $m \ge 2$, $G'_j$ has fewer vertices. Also, the maximum degree of $G'_j$ cannot exceed $\Delta_G$. Thus, we can find a $\Delta_G$-colouring of $G'_j$ by the inductive hypothesis unless $G'_j$ is a complete graph or an odd cycle. But even in these cases, we have
$$\Delta_{G'_j} = \deg_{G'_j} A \le \deg_G A - 1 \le \Delta_G - 1.$$
This shows we can still find a colouring using at most $\Delta_G - 1 + 1 = \Delta_G$ colours.

Now, we have found a $\Delta_G$-colouring for each $G'_j$. WLOG assume $A$ has colour 1 in all such colourings. Then we can combine the colourings used for all $G'_j$ to give a colouring for $G$. The condition is satisfied, since any two of $G_1, G_2, \dots, G_m$ are disconnected.

Case 2. $G - \{A\}$ is connected for all vertex $A$, and there exist two non-adjacent vertices $B$ and $C$ such that $G - \{B, C\}$ is disconnected.

Suppose the connected components of $G - \{B, C\}$ are $G_1, G_2, \dots, G_m$. Note that $B$ must be adjacent to some vertex in each $G_j$, or otherwise we can remove $C$ to get at least two connected components (one is $G_j$), contradicting the assumption. The same holds for $C$. For each $G_j$, let $H_j$ be the induced subgraph of $G$ containing $B, C$ and all vertices in $G_j$.
Now, the degrees of $B$ and $C$ in $H_j$ are at most $\Delta_G - 1$ since $m \ge 2$. Consider the graph $H'_j$ obtained from $H_j$ by including the edge $B-C$. Then we still have $\Delta_{H'_j} \le \Delta_G$. So there is a $\Delta_G$-colouring for $H'_j$ unless it is an exceptional case. If such a colouring exists for all $H'_j$, WLOG we may assume $B$ has colour 1 and $C$ has colour 2 in all colourings (note that their colours are different since $B-C$ is an edge). Then we can combine all the colourings to give a $\Delta_G$-colouring for $G$.

Note that the same argument works as long as there is a $\Delta_G$ colouring for each $H'_j$. In particular, the argument works if some $H'_j$ is an odd cycle (since $\Delta_G \ge 3$) or a complete graph with at most $\Delta_G$ vertices. So the only case left is that $H'_1$, for example, is a complete graph with $\Delta_G + 1$ vertices. Since the degrees of $B$ and $C$ in $G$ are at most $\Delta_G$, each of $B$ and $C$ can only be adjacent to one more vertex not in $G_1$. Therefore, we have $m=2$, and each of $B$ and $C$ is adjacent to only one vertex in $G_2$.
Now, we can identify $B$ and $C$ in $H_2$ as the same point $D$ to form a new connected graph $H'$. Since $\deg D \le 2$ and the degrees of all other vertices are unchanged, we have $\Delta_{H'} \le \Delta_G$ and $H'$ has fewer vertices than $G$. Therefore, there is a $\Delta_G$-colouring of $H'$. This also holds even if $H'$ is an odd cycle or a complete graph (which has at most 3 vertices as $\deg D \le 2$). This yields a $\Delta_G$-colouring of $H_2$ such that $B$ and $C$ have the same colour (as $D$). Then we can colour the remaining $\Delta_G - 1$ vertices in $G_1$ using colours different from that of $B$ and $C$. This gives a $\Delta_G$-colouring of $G$.

Let $A$ be a vertex with $\deg A = \Delta_G$. If all neighbours of $A$ are adjacent, then there is already a $(\Delta_G + 1)$-clique. This is impossible since the graph is connected and not complete, but we cannot add more edges to this clique (or $\Delta_G$ is increased). Suppose $B$ and $C$ are some non-adjacent neighbours of $A$.
Starting from $A_1 = A$, we construct a sequence of vertices as follows. Suppose we have constructed $A_1, A_2, \dots, A_n$. Since $G - \{B, C\}$ is connected, we can always find some $A_{n+1} \neq B, C$ such that $A_{n+1}$ is adjacent to one of $A_1, A_2, \dots, A_n$ unless all vertices are used. We repeat the same process until we have obtained $A_1, A_2, \dots, A_{k-2}$. Then we define $A_{k-1} = B$ and $A_k = C$.

We colour the vertices from $A_k$ back to $A_1$ as follows. We can assign colour 1 to $A_k$ and $A_{k-1}$ since they are non-adjacent. For each $A_j$ we can always assign a colour to it which is different from its neighbours in $A_k, A_{k-1}, \dots, A_{j+1}$ since $\deg A_j \le \Delta_G$ and one of the neighbours of $A_j$ is $A_1, A_2, \dots, A_{j-1}$ by construction. We can do this for $A_{k-2}, A_{k-3}, \dots, A_2$. Lastly, for $A_1$, it has $\Delta_G$ neighbours, where 2 of them ($A_{k-1}$ and $A_k$) have the same colour. So we can still find a new colour for $A_1$. This yields a $\Delta_G$-colouring for $G$.

The proof is complete by combining all these cases and using induction.

e.
An example is a complete graph with $k$ vertices. By part (c), we have $\chi(G) = k$. Also, $\overline{G}$ has no edges, and so $\chi(\overline{G}) = 1$. This gives
$$\chi(G) + \chi(\overline{G}) = k + 1$$