Probability when arranging $N$ different numbers in a line we would like to arrange the numbers: $ 1, 2, \ldots , N$ in a line. 


what is the probability that digits $1, 2$ are not arranged next to each other?
what is the probability that digits $1, 2$ are not  arranged next to each other, and also the digits $3, 4$ are arranged next to each other?


my attempt:

This is mostly a combinatorics question, as the "Probability", meaning: $$\Omega = N!$$


I treated $1, 2$ as a unit, looking for the number of different ways to arrange them next to each other - which would be the Complementary event.
$$P(A)^C = \frac{(N-1)! \cdot 2 }{N!}$$
where there's in fact $N-1$ needed to be arranged, ans 2 is the number of permutations.


so that $$P(A) = 1- \frac{(N-1)! \cdot 2 }{N!} = 1 - \frac{2}{N}$$

for the second sections - I'm not sure.
suppose I'd be looking the number of different ways to arrange  $1, 2, \ldots N$ in a line so that the digits $1, 2$ and $ 3, 4$ are always arranged next to each other, then that will be:
$$\frac{(N-2)! \cdot 2 \cdot 2}{N!}$$ where 4 is the number of permutations (2 for each unit of digits)
but does that mean that there's $$\left(N! - {(N-2)! \cdot 2 \cdot 2}\right) \cdot \left((N-2)! \cdot 2 \right)$$ ways to arrange the numbers? 

that doesn't seem right.
The number of ways that $1$ and $2$ appear next to each other can be calculated by considering both the numbers as a single unit. So the packets will be $$ (1,2),(3),(4),(5) ....... (N) $$ 
Number of ways these can be arranged are $(N-1)!$ as there are $(N-1)$ packets. To consider the backward and forward arrangement of 1 and 2 inside the packet we need to multiply by two. So,
Total ways for 1 and 2 to be arranged together is $2(N-1)!$  

So the numerator will be (Total ways-Undesired ways).  $$N! - 2(N-1)!$$.
Divide this with the total ways we get $$\frac{N! - 2(N-1)!}{N!}  = 1 - \frac 2N$$  

For the second part we consider one packet with 1 and 2. The other packet with 3 and 4. So the packets will be,$$ (1,2),(3,4),(5),(6),(7) ....... (N) $$ 

Arrangements in which 1 and 2 are together(like the first part) are $P(X)=2.(N-1)!$
Arrangements in which 3 and 4 are together are $P(Y)=2.(N-1)!$
Arrangements in which both groups (1 and 2) and (3 and 4) are together are $P(Z)=2.2.(N-2)!$. The two packets have an internal arrangement of two ways and the total count of packets is $(N-2)!$ 

The numerator in the second part is $N! - P(X) - P(Y) + P(Z)$.
The denominator will be $N!$
Regardless of the other numbers in the list, there are $2(N-1)$ ways to position $1$ and $2$ next to each other in a list of size $N$. Assume they take the first position, then you have to sort the other $N-2$ values into a list of size $N-2$. Thus we have that the total ways of satisfying the question is:
$$2(N-1)\cdot (N-2)!=2(N-1)! $$

The probability follows nicely from this, it is $$\frac{2(N-1)!}{N!}=\frac 2N $$

This approach should also work for (b).