checking whether turing machine passes at least k>2 states before accepting a word
Clearly $L$ is acceptable (just simulate $M$ and keep track of the number of distinct states encountered during the simulation). We now show that it is not decidable.
If $L$ were decidable, you would be able to solve the Halting problem as follows:
given a TM $T$ and an input $x \in \Sigma^*$, construct a TM $M$ that ignores its input, simulates $T$ with input $x$ and, when the simulation is complete, accepts.
You can further ensure that, if $M$ accepts, then it also traverses at least $3$ distinct states by just transitioning from the initial state to another (distinct) state before starting the simulation of $T$.
Now check whether $\langle M, 3 \rangle \in L$. If the answer is affirmative, then there is some $w \in \Sigma^*$ for which $M(w)$ accepts, showing that $T(x)$ halts.
If the answer is negative then $M$ never halts, showing that $T(x)$ does not halt.