Convergence in norm but not almost everywhere of $f_{t_n}=f(\cdot-t_n)$ to $f$

Your idea of a fat Cantor set works too. I'll repeat a construction I used in this answer.

Let $C$ be a compact nowhere dense set of positive measure. We can construct a sequence $t_n \to 0$ such that for every $x \in C$, there are infinitely many $n$ for which $x - t_n \notin C$. In particular we have $\liminf_{n \to \infty} \chi_C(x-t_n) = 0$ and therefore $\chi_C(x-t_n) \not\to 1 = \chi_C(x)$. Since $C$ has positive measure, we therefore do not have $\chi_C(\cdot-t_n) \to \chi_C$ a.e.

To construct the sequence, let $U = C^c$ and note that $U$ is open and dense. So for each integer $m > 0$ and each $x \in C$, there exists $u \in U$ such that $|x-u| < 1/m$, which is to say that $x \in U+(x-u)$. In other words, the collection of open sets $\{U+s : |s| < 1/m\}$ cover $C$ (indeed they cover $\mathbb{R}$). Since $C$ is compact, there is a finite subcover $\{U+s_{m,1},\dots, U+s_{m, k_m}\}$. That is to say, for every $x \in C$ there is some $1 \le i \le k_m$ for which $x - s_{m, i} \in U = C^c$.

Now let $(t_n)$ be the sequence $$(s_{1,1}, \dots, s_{1, k_1}, s_{2,1}, \dots, s_{2, k_2}, \dots).$$ Since $|s_{i,m}| \le 1/m$ for every $m$, we have $t_n \to 0$, and there are infinitely many $t_n$ such that $x-t_n \notin C$ (namely, at least one of the $s_{m,i}$ for each $m$).

Yes. Consider $f(x)=\sum 2^{-n}g(x-q_n)$, where $q_n$ is an enumeration of $\mathbb Q$, and $g\in L^1$ is an unbounded non-negative bump function such as $g(x)=\chi_{(-1,1)}(x)|x|^{-1/2}$.

I'm now going to produce a sequence $t_n\to 0$ for which $f(x-t_n)$ is unbounded for every $x\in [-1,1]$. My first bunch of $t_n$'s will satisfy $|t_n|\le 1$. Start out by locating a $q_n$ close to $-1$ ($q_n=-1$ would work nicely). The corresponding bump $2^{-n}g(x-q_n)$ will be $\ge 1$, say, on an interval centered at $q_n$. By shifting this bump around, I can make sure that $f(x-t_n)\ge 1$ for every $x\in [-1,0]$ for some $n$. Cover the remainder of $[-1,1]$ in the same way.

Let's summarize: I have chosen finitely many elements $t_1, \ldots , t_N$ of my sequence, in such a way that $f(x-t_n)\ge 1$ for every $x\in [-1,1]$ for some $1\le n\le N$. Moreover, these satisfy $|t_n|\le 1$.

Now just continue in this way: In the second step, find the next batch of $t$'s, such that $f(x-t_n)\ge 2$ for all $x\in [-1,1]$ for some $n$, and $|t_n|\le 1/2$ for these $t_n$ etc.