Suppose $C$ is a nonsingular plane projective cubic. This exercise starts by assuming that $O$ is a flex, and first asks to prove that the points of $C$ of order $3$ in $C$ are precisely the flexes other than $O$, and that together with $O$ they form a subgroup isomorphic to $\mathbb{Z}/(3)\times \mathbb{Z}/(3)$. This is not hard.
It then goes on to ask for a proof that the points of order 2 (together with $O$) form a subgroup isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$. This is also not hard to show.
So, First question: From that, can we conclude (since the structure of the group is independent of the basepoint) that no matter which basepoint we choose (whether it is a flex or not) there are three points of order 2 which, together with the basepoint, form a Klein 4-group? (Of course, the identity of these points changes when the basepoint changes.)
The final part of the exercise asks: "Let $C$ be a nonsingular cubic, $P\in C$. How many lines through $P$ are tangent to $C$ at some point $Q\ne P$? (The answer depends on whether $P$ is a flex.)" It's the final parenthetical comment that I don't understand. If my first question is correct, then given a point $P$, set $O=P$. There are then exactly three lines from $P$ that are tangent to a point $Q$ other than $P$: the lines through the points of order 2. So Second question: Why is the answer dependent on whether $P$ is a flex or not? (I understand DCT's comment, and it makes sense [assuming you can find one such $Q$], but I don't see where my arguments fail.)
Edited to respond to DCT's comment below and to clarify my confusion.
This document of mine will hopefully answer 95% of your question: https://www.overleaf.com/read/kgyxszthnsxp.