For a constructive proof on (1):
First consider a node $v$ that has degree $ \ge n/2$. Then call $W$ the set of nodes that are not directly connected to $v$ noting that $|W|\le n-n/2-1 = n/2-1$. Then consider any node $w \in W$ and note that it must be connected to some node outside $W$, i.e some node that is connected to $v$. Thus the graph is connected.
We can reduce the minimum degree value here to $\ge \lfloor n/2\rfloor$, but smaller values allow the separation of the graph into two complete graphs each with half the nodes.
For question (2) a path length of $>n/2$ between chosen nodes $u$ and $v$ indicates that there is a node on that path that can be removed to break the path and hence disconnect the graph, but I can't quite put the argument together with the rigor I would like.
Certainly though the limit cannot be much reduced as the cycle graph with even nodes has $2$ paths of length $n/2$ between any two points, so there is no single node can be removed to disconnect the graph.