Let $X=(X_n)_{n\geq 0}$ be a symmetric random walk on $\mathbb{Z}$ with $X_0=0$. I have to prove/disprove that $\tau_1=\inf \{n \in \mathbb{N}_0 : n\geq 11, \, X_n\cdot X_{n-10}\leq 5\}$ and $\tau_2=\sup \{n \in \mathbb{N}_0 : X_n>42\}$ are stopping times with respect to the canonical filtration.
I think that $\tau_1$ is indeed a stopping time whil $\tau_2$ is obviously not a stopping time since we don't know what $n$ is. For $\tau_1$, I have to show that $\{\tau_1\leq n\}\in \mathcal{F}_n$ for all $n$. I believe I have to write this set in terms of unions/intersections/complements of sets I know are in $\mathcal{F}_n$. But I fail to do so. For $\tau_2$ I suppose I have to construct a counter example? But how do I do that in this case?
I give you an idea for $\tau_1$ : by definition, $\tau_1<T$ if and only if for all $10\le n\le T-1$, we have $X_n X_{n-10}> 5$.
Now for each $n$, you can "condition" on the value of $X_n$, since you know it is $\mathbb Z$-valued. In other words, if $X_n =j\in \mathbb N^+$, $X_n X_{n-10}> 5$ holds if and only if $X_{n-10}> 5/j $ (note that $j$ can not be $0$ on that event, do you see why ?). Similarly, if $X_n=-j$, then the condition holds if and only $X_{n-10}< -5/j $.
Formalizing these observations, we can thus write $$\{\tau_1 < T\} = \bigcap_{10\le n\le T-1}\bigcup_{j\in\mathbb N^+\\ j\ne 0}\left\{X_n=j,X_{n-10}>5/j\right\} \cup \left\{X_n=-j,X_{n-10}<-5/j\right\} $$ Since $\{\tau_1 < T\}$ is made of countable unions and intersections of elements of $\mathcal F_T$ it follows that it is an element of $\mathcal F_T$ itself.
If you understood this, you should be able to make similar construction and prove that $\{\tau_1 \le T\}\in\mathcal F_T$.
For $\tau_2$, you(ve got the correct intuition : "we don't know what $n$ is". To make it formal, just write down the event $$ \{\tau_2 = T\}=\{X_T > 42\}\cap\bigcap_{n>T} \{X_n\le 42\}$$ Since $\{X_n\le 42\}\not\in \mathcal F_T$ for any $n>T$, it follows that $\tau_2$ is not a stopping time.