Here is a way of visualizing the situation, in your case of $n=5$.
There are $5$ chairs, evenly spaced around a circular table. These chairs are labelled $1,2,3,4,5$, say counterclockwise.
We want to seat $5$ people $A,B,C,D,E$, so that seating arrangements that differ by a rotation are considered the same. The string $BAEDC$ means $B$ is at $1$, $A$ at $2$, $E$ at $3$, $D$ at $4$, and $C$ at $5$. This is considered the same as $AEDCB$, also $EDCBA$, $DCBAE$, and $CBAEDE$.
Call these $5$ arrangements a family. Any arrangement belongs to a unique family of $5$.
There are $5!$ arrangements of the letters $A,B,C, D, E$ in a line. We ask how many families there are. Well, since each family contains $5$ strings, there are $\dfrac{5!}{5}=4!$ families. Thus there are $4!$ circular permutations of our $5$ people.
With $n$ people, exactly the same idea works, and there are $\dfrac{n!}{n}=(n-1)!$ circular permutations.
Here we considered two arrangements, one of which is clockwise, and the other anti-clockwise, as different.
But if we consider two such arrangements as being the same, then for example $ABCDE$ and $EDCBA$ are considered the same. This means (in our case) that now each family has $10$ objects, twice as many as before. So the number of "really different" arrangements in this case is $\dfrac{5!}{(2)(5)}$.
This generalizes nicely to $n$ people, with a couple of minor exceptions.
If $n\ge 3$, the number of circular arrangements with clockwise and anti-clockwise considered the same is $\dfrac{n!}{2n}=\dfrac{(n-1)!}{2}$.
If $n=1$, obviously there is only $1$ arrangement, whatever the conditions. And if $n=2$, the one and only arrangement is the same, whether clockwise or anticlockwise are considered the same or not.
For 1, without the condition you can select $5 \choose 3$ sets. If $m$ has to be included, you can select $4\choose 2$ for the other letters. When you talk of sequences, you care about order. The easiest way for this is to calculate the total number of sequences, then subtract the ones that don't have $m$ in them somewhere. Choosing a sequence of four items out of five, you have five choices for the first, four for the next, etc. so there are 5*4*3*2=120. Without $m$ you have 4*3*2*1=24,so the ones that have $m$ are $96$ Your last is English, not Math, but yes "at most once" means you cannot repeat.
Best Answer
In this small problem the easy way is to note first that there are $4\cdot3\cdot2=24$ ways to pick three different letters and permute them. Then observe that there are $3$ ways to pair LL with one of the other letters, and $3$ ways to place that odd letter relative to the two L’s, so there are $3\cdot3=9$ more $3$-letter arrangements that include both L’s, for a total of $24+9=33$ arrangements.