I'm struggling with two different-but similar questions. I'm pretty new with the subject so will really appreciate explanation on how to approach these kind of question
Prove that in every 10 vertices graph with 28 edges there are 2 vertices with two different neighbors.
Prove that in any graph G with $|V|=n$ and $d(v)=\sqrt(n)$ $\forall v\in\ V$ there is a triangle or 2 vertices with two different neighbors.
The title talks about "common neighbours", but the question itself is about "different neighbours". This answer will convince you that "common neighbours" are meant.
We need a few lemma's. I'll just state them, they are easy to prove, let me know if you have trouble. $G$ is our graph, $n$ is the vertex count, $e$ the edge count.
Lemma 1: The average "sum of degrees" over pairs of different vertices is twice the average degree (in formula: $\frac{\sum_{v\ne w}(d(v)+d(w)}{n}=\frac{4e}{n}$).
Lemma 2: If we have vertices $v$ and $w$ with $d(v)+d(w)\geq n+2$ then $v$ and $w$ have at least two common neighbours.
Lemma 3: If we have vertices $v$ and $w$ with $d(v)+d(w)\geq n$ and $G$ is triangle-free then $v$ and $w$ have at least two common neighbours.
The first problem: Lemma 1 tells us that the average "sum of degrees" is $\frac{4e}{n}>11$. The pigeonhole principle now gives us a pair $v,w, v\ne w$ with $d(v)+d(w)\geq 12=n+2$, so lemma 2 tells us that $v$ and $w$ have two common neighbours.
Note that the proof fails for $e=27$, which is strong evidence that "common neighbours" is the proper choice to make.
The second problem is easier (or at least more common knowledge). Note that "no two vertices with 2 common neighbours" is equivalent to "no $C_4$". So we just need to show that a graph with girth 5 (no triangle $C_3$ and no $C_4$) and minimum degree degree at least $n$ has more than $n^2$ vertices and this is a standard result:
let $v$ be an arbitrary vertex. $v$ has at least $n$ neighbours $w_1,\ldots,w_n$. Each of this neighbours has itself has at least $n-1$ neighbours different from $v$. $w_i$ cannot be a neighbour of $w_j$ (or we would have triangle $vw_iw_j$) and $w_i$ can $w_j$ cannot have a common neighbour $z$ (or we would have $C_4$ $vw_izw_j$). So $G$ has at least $1+n+n(n-1)=n^2+1$ vertices.