Knapsack on two kinds of objects, where you cannot choose type 2 objects on their own
I think this can be solved the same way that knapsack problem is solved using dynamic programming.
I present the problem for the case where type 1 items are to be selected in pairs, but the other case is solvable in a similar way.
Suppose the weights are $\{w_1, …, w_n\}$, the profits are $\{p_1, …, p_n\}$ and the types are $\{t_1, …, t_n\}$. Let's define, for $w\in [\![0, W]\!], i \in [\![1, n]\!], j\in\{0,1\}$, $K(w, i, j)$ representing the maximum profit using a knapsack of capacity $w$, using items $1, …, i$, with a parity of type 1 items equal to $j$.
We want to determine $K(W, n, 0)$. Using this definition, we can calculate $K(w, i, j)$ by distinguishing if we take or not the $i$-th item:
$K(w, i, j) = \max(K(w-w_i, i - 1, j \oplus t_i) + p_i, K(w, i - 1, j))$ (where $\oplus$ is the addition modulo 2)
With this formula, we can then compute $K(W, n, 0)$ in time complexity $O(nW)$.
In the case you just need to pick at least two items of type 1 if you want any, $K(w, i, j)$ could represent the maximum profit with $j\in \{0, 1,2\}$ representing respectively exactly zero type 1 item, exactly one type 1 item or at least 2 type 1 items.