The graph $G=(V, E)$ is given by the following diagram. Solve the tasks for the graph. 
For which maximum $k$ in $N$ is $G$ $k$-fold connected? Give reasons for your answer! Give the blocks of $G$ by a diagram each. Give for the block graph $G_B$ of $G$, its node set %V_B% and its edge set $E_B$ and draw a diagram of this block graph. Explain why at least three edges have to be added to $G$ (while maintaining the node set) so that the block graph of the resulting graph consists of exactly one isolated point. Give three edges that accomplish this.
Problem/Approach:
For the first task, I have that it is $1$-fold connected because if it was $2$-fold connected, if I remove the edge between $1$ and $3$, node $3$ is completely without an edge connection, and therefore it would not be $3$-fold connected.
For task 2, I have the blocks:
Block $1$: ${1,3}$
Block $2$: ${1,2,4,6}$
Block $3$: ${6,5}$
Block $4$: $(6,7)$
Block $5$: ${6,8,10,11}$
Block $6$: ${11,9}$
$1$, $6$ and $11$ are the hinge points.
$V={B1,B2,B3,B4,B5,B6,1,6,11}$
$E_B = {{B1,1},{1,B2},{B2,6},{6,B3},{6,B4},{6,B5},{B5,11},{11,B6}}$
For task 3, the reasoning is that each block is a maximum connected part of the graph. To reduce the block graph to one isolated point, each node must be in its own block. This means that each node must only be connected to itself, and there can be no edges between nodes.
To achieve this, at least three edges have to be added, as each node has at least one edge.
The at least three edges that can be added are for example: $(1, 1), (2, 2), (3, 3)$ or $(1, 1), (2, 2), (4, 4)$ or $(5, 5), (6, 6), (7, 7)$
These edges create a kind of "loop" where each node is connected to itself and no edges between nodes exist. The block graph of G would then consist of exactly one isolated point.
Could someone help me to see if I solved tasks 1 and 2 correctly, if not then if someone could write his solution for me. Could someone tell me in task 3 if the idea is correct, if not then if someone could write his idea for me. I would be really happy if you could help me.
 A: I follow this instruction.
To my understanding, we consider vertex connectivity, not edge connectivity.
So, the first answer is $k=1$, and the cut vertices are $1,6,11$.
For the second question on the blocks, we proceed as follows. First, we remove and duplicate $6$, which gives the induced subgraphs $G_1$ induced by $1,2,3,4,6$, further $G_2$ induced by $5,6$, then $G_3$ induced by $6,7$, and finally $G_4$ induced by $6,8,9,10,11$. For $G_1$ we also have the cut vertex $1$, which gives the subgraphs $G_{11}$ induced by $1,3$ and $G_{12}$ induced by $1,2,4,6$. Similarly, for $G_4$ we get $G_{41}$ induced by $6,8,10,11$ and $G_{42}$ induced by $9,11$. Now, the blocks are $B_1=G_{11}$, $B_2=G_{12}$, $B_3=G_2$, $B_4=G_3$, $B_5=G_{41}$ and $B_6=G_{42}$.
For the third question on the block graph, we have the (bipartite) graph $B=(V,E)$, where the vertices $V=\{1,6,11\}\cup\{B_i:1\le i\le 6\}$ are the cut vertices and the blocks, and the edges are $E=\{\{B_1,1\},\{B_2,1\},\{B_2,6\},\{B_3,6\},\{B_4,6\},\{B_5,6\},\{B_5,11\},\{B_6,11\}\}$.
For the fourth part of the question, we notice that $3,5,7,9$ have to be connected by an additional edge, since the graph has to be $2$-connected for the block graph to be trivial, and hence in particular the minimum degree has to be $2$. However, for any of these three possibilities to add two edges (add $35$, $79$ or $37$, $59$ or $39$, $57$), the vertex $6$ is still a cut vertex. Hence we need to add at least three edges. But adding $25$, $37$ and $59$ yields a $2$-connected graph. This also answers the last question.
