Here is a start, sort of. The problem is that I cannot think of a way of doing it in my preferred way without using the equality symbol $\:=$. (So I am working in the predicate calculus with equality.)
$1$. We want to say roughly that there is an $x$ such that for all $y$, $x$ and $y$ are joined by an edge. But more precisely, we probably don't quite want to say this, for we probably do not want to say that the $x$ is joined to itself. That complicates things slightly. Ask yourself whether the following does the job:
$$\exists x\forall y(\lnot(x=y)\implies G(x,y)).$$
Equivalently, we can use $\exists x\forall y((x=y)\lor G(x,y))$
But this allows for the possibility that the point $x$ is connected by an edge to itself. If we want to rule that out explicitly, we can use $\exists x(\lnot G(x,x)\land \forall y(\lnot(x=y)\implies G(x,y)))$
$2$. The idea is quite similar to the one in Problem $1$, except that we have to say that for all $x$ and $y$, if $x\ne y$ then $x$ and $y$ are joined by an edge. Here I am interpreting "any two vertices" as meaning any two distinct vertices.
$3$. We need to interpret the meaning of the informal sentence. Is it any two vertices or is it any two distinct vertices? And do we allow the possibility of loops? I will assume no loop is being used, and that the two vertices we want to say are linkable are distinct. So we want to say that for any $x$ and $y$, if $x\ne y$, there exist objects (vertices) $u$ and $v$ such that $x\ne u$, and $x$ and $u$ are joined by an edge, and $u\ne v$, and $u$ and $v$ are joined by an edge, and $v\ne y$, and $v$ and $y$ are joined by an edge. The sentence will be kind of long, but not terribly complicated in structure.
$4.$ We want to say that for all vertices $x$, $y$, and $z$, these vertices do not form a triangle. You will have to decide whether your sentence allows loops, that is, whether $G(x,x)$ is allowed.
$5$. We want to say that for any $w$, $x$, $y$, $z$, presumably distinct, $w$ and $x$ are joined by an edge, or $w$ and $y$ are, or $w$ and $z$ are or $\dots$.
Comment: If we want to interpret $1$ as saying that there is a vertex connected to everybody, including itself, things are much easier. We can simply write $\exists x\forall y G(x,y)$. But that is probably not what a graph-theorist intends.