Not a full answer. Actually, with the reference included, it pretty much is.
Here is only the first part for the triangle.
Suppose that we have a critical graph $G$ which is not triangle free. Consider any triangle within such a graph. $G$ cannot be the triangle graph itself, so at least one vertex has a component $C_1$ connected to it (with possibly multiple edges). Let's say $v_1$ connects to $C_1$, $v_2$ to $C_2$ and so on.
I claim that each component is connected to either
a) another vertex of the triangle
b) a component connected to another vertex of the triangle
because otherwise the removal of the corresponding vertex will disconnect the component so the graph is not biconnected. But then if $C_1$ is connected to $v_2$ or a component of $v_2$ the removal of $e = (v_1, v_2)$ will not introduce any articulation point. This is a contradiction. To see this a bit more clearly, consult the example in the image below. Either the blue dotted edge must be present or both red dotted edges must be present.
Edit I may have worked something out for the third part. I will include a sketch with details for you to fill in.
Let $G$ be a critical graph. Then $G$ is biconnected and in particular this implies that there exists a cycle between any two vertices (via Dirac). Suppose for the sake of contradiction that there exists a critical graph which has no vertices of degree $2$.
Lemma: If $C$ is a cycle in a critical graph $G$ then there must not be any edges joining non-adjacent vertices of $C$.
Proof: If there exists any such edge, then removing it still leaves the graph biconnected and hence the graph is not critical.
Here is a sketch then
- Since there are no vertices of degree $2$, any cycle of $G$ must be part of a nested cycle.
Consider a nested cycle of two layers (shown above) $C$. I've labelled the exterior vertices from $v_1$ to $v_n$ and the interior from $u_1$ to $u_m$. I've labelled edges as solid lines and paths as dotted lines. In the path nested within the exterior cycle, $u_1$ must have another neighbor since it is of degree greater than $2$. Call such a neighbor $w$.
$w$ must not be any vertex on the nested cycle itself otherwise the edge $(u_1, w)$ violates the above lemma. Therefore $w$ is distinct from $C$. There is a cycle containing $w$ and a vertex of the exterior cycle. In particular, there exists at least one path connecting $w$ to the exterior cycle. Let $P$ denote the shortest such path from $w$ to some $v_i$.
- The existence of $P$ creates another cycle (in red)
$$(w, P, v_i, \cdots , v_n , v_1 , \cdots, v_k, u_m, \cdots , u_1, w)$$
but edge $(v_1, u_1$ connects non-adjacent vertices of the above cycle and hence the graph cannot be critical.
The above should be enough to prove that there exists a vertex of degree $2$ provided that I made no errors. I will leave you to check through the argument. Of course feel free to ask for clarification.
Edit I found this very nice book which outlines all of the above properties. The book calls your critical graphs as "minimally $2$-connected graphs". The proof for the triangle property is essentially as I outlined, but with much more clarity. In particular there is also a proof of the subgraph property and the vertex of degree two property in the book (actually they prove something significantly stronger). If you are interested in problems like these, you may want to check out the general theory of extremal graph theory.