Team construction in tri-partite graph
An upper bound on the first question is $k\leq 15$: Take a set of $5$ $a$s $A=\{a_1,\dots,a_5\}$, $5$ $b$s $B_1=\{b_1,\dots,b_5\}$ and $5$ $c$s $C_1=\{c_1,\dots,c_5\}$. We know that at most 2 of the $a$s don't have a neighbor among the $B$s, otherwise we found a complement of a $K_{3,3}$, which is forbidden. The same holds for $a$s and $c$s. Thus there must be one $a_1'$ that has a neighbor among both sets. We call these neighbors $b_1'$ and $c_1'$ respectively.

We now fix the set $A$ and consider $10$ additional pairs of sets of $b$s and $c$s $(B_i,C_i)_{i\in\{2,\dots,11\}}$ with 
$$B_i=\{b_{i+4}\}\cup(B_{i-1}\setminus \{b_{i-1}'\})$$ and 
$$C_i=\{c_{i+4}\}\cup(C_{i-1}\setminus \{c_{i-1}'\})$$
and choose $b_i'$ and $c_i'$ such that they're both neighbors of the same $a_i' \in A$ (which all exist by the observation above). 

Now at least $3$  pairs of sets agree on the same $a$ by the pigeonhole principle, i.e. there is an $a_l\in A$ and pairwise different $m_1,m_2,m_3 \in \{1,\dots,11\}$ such that $a_l = a_{m_1}' = a_{m_2}' = a_{m_3}'$.
Now $b_{m_p}'$ and $c_{m_p}'$ are neighbors of $a_l$ for $p\in\{1,\dots,3\}$. Thus for some $p,p'\in \{1,2,3\}$ the set $\{a_l,b_{m_p}',c_{m_{p'}}'\}$ induces a triangle of friends.