I can prove a lower bound of
$69$ tries.
The first step is to notice that the question is equivalent to
finding an $132$-node graph $G$ with no independent set of size $\ge 66$.
The reason is:
1. If we have the graph $G$, then for each edge $(x,y)$ in $G$, we put the $x$-th and $y$-th card into the slot. If we failed then no edge has two green cards as its endpoint, so $66$ green cards induce an independent set of $G$.
and
2. If we have a strategy to win the game, then the strategy must be nonadaptive, since we don't gain any information from each failure to choose two green cards. That is, the strategy of each step is already determined before the game gets played. If the $i$-th step is to test card $a_i$ and card $b_i$, we insert edge $(a_i,b_i)$ into $G$. If $G$ has a size-$66$ independent set, our strategy would be invalid when this set of cards are green.
Then we observe
$G$ must contain an odd cycle. Because otherwise $G$ is a bipartite graph which always have an independent set of size $\ge \frac{132}{2}=66$.
and
If $G$ has $C$ connected components, then $G$ has at least $133-C$ edges. For an arbitrary graph $G'$, if it has $C'$ components and $V'$ vertices, then it has at least $V'-C'$ edges, so we infer that $G$ has at least $132-C$ edges. However as we demonstrated above, $G$ has a cycle, and deleting an edge from that cycle doesn't change the number of its components. So we conclude $G$ has $\ge (132-C)+1$ edges.
So there are a few cases. If
$C\ge 66$, then we can take any vertex from a component to form an independent set of size $C$. This is invalid;
If
$C\le 64$, then $G$ has $\ge 133-64=69$ edges;
If
$C=65$, then every component must be a clique: if some component is not a clique, we can take two vertices from it and one vertex from every other component and still form an independent set (of size $C+1$). Now let $a_1,a_2,\dots,a_{65}$ be the number of vertices in each component, then $\sum_{i=1}^{65}a_i=132$ and the number of edges is $E=\sum_{i=1}^{65}\frac{a_i(a_i-1)}{2}$. Even if we allow $a_i$'s be real numbers (instead of integers), the minimum of $E$ is reached when each $a_i=a=132/65$, and $E=65\cdot \frac{a(a-1)}{2}\approx 68.03$. Therefore $69$ edges are needed in this case. (We note that this corresponds to @trolley813's answer, and optimal integer solution is that each $a_i$ is $2$ except two of them are $3$.)
