My friend asked me this question. First I am wondering whether this problem is well defined. However, my attemp was:
a. Counting all 4 digits numbers from $\{1,2,3,4,5\}$ which is $5 \cdot 4 \cdot 3 \cdot 2= 120$ options. I meant that no double $1$ will appear.
b. $4$ digits numbers with double $1$: there are $4$ ways to locate $8$ twice. multiplying by $4 \cdot 3=12$ options to fill in the last two digits we get: $48$ options.
So eventually there are $168$ numbers that satisfy these conditions.
Am I correct? should it be written more rigorously?
Take the complement of your second and third cases. $-11-$ and $-1-1$