Decidability of the language of all deterministic LBA where all states are reachable

Question

I have a exam task with 3 parts. (b) is no problem. I got a solution for (a) but the way (c) is asked makes me wonder if I even understood (a)

(a) L := { < A > | A is a DFSA, where all states are reachable }. Is L decidable? Why or why not?

(b) M is a DLBA and k is the input length. There exists a function g(M,k) which returns the maximum number of steps a DLBA can possibly do, which halts on an input the size of k. Why?

(c) L' := { < M > | M is a DLBA, where all states are reachable }. Show that L' is semi-decidable. Use g(M,k) from (b)

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My solution:

(b) Limited amount of configurations. I get that.

For example how I understand (a): L is a language and every word of L is the encoding of a FSA. But only FSAs where every state can somehow be reached from the starting state or is the starting state.

So is it decidable?

My answer: Yes, for any word w you could check if its a valid FSA-encoding, if not: reject, if it is: Construct a FSA from this encoding and check if you can reach all states like this:

Mark the starting state;
Repeat until no new marked states:
   Mark all states you can reach from the marked states.
Check if all states are marked. 
Reject or Accept

My Question:

Is (c) just semi-decidable and how do I use g(M,k) to show this? Because I would think I could solve (c) exactly like (a) and show that it is decidable (which would mean it is also semi-decidable)

Edit: Hi, if you give me a downvote I would appreciate a comment on how to improve my questioning. I know it is a long post but I just try to lay out how I understand questions like this to be solved so you can correct me if I have key misunderstandings. I now reduced the amount of questions I asked and the length of my post.