In these problems, a "tree diagram" is useful, indeed almost indispensable, for analysis of the situation. Tree diagrams have been left out, because they take some time to draw with graphing software. So we are left with purely "verbal" arguments. Please, in each case, do draw an appropriate tree diagram!
Problem 1:
Way 1: The probability that the sum on any one toss is $5$ is $4/36$. The probability that the sum is $7$ is $6/36$. Maybe it is obvious that with probability $1$ "first is $5$" or "first is $7$" will happen. And maybe it is obvious that the ratio of the probabilities of the two events is $4$ to $6$. Then the probability that the first is $5$ is $4/10$. This turns out to be correct, but unless one's intuition is very well developed, this kind of reasoning can be dangerous.
Way 2: When we toss two dice, let $N$ be the event "we get neither a $5$ nor a $7$." It is easy to see that the probability of $N$ is $26/36$.
Imagine the game is over when either a $5$ or a $7$ happens. Then "$5$ comes before $7$" can happen in several ways. We could get a $5$ immediately. Call that event $5$, Or we could get $N$ then $5$. Call this $N5$. Or we could get $NN5$. Or else we could get $NNN5$, and so on.
The probability of $5$ is $4/36$. The probability of $N5$ is $(26/36)(4/36)$. The probability of $NN5$ is $(26/36)^2(5/36)$. And so on.
So the probability of "$5$ before $7$" is
$$\left(\frac{4}{36}\right) + \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)+ \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)^2 + \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)^3 +\cdots$$
Sum the above infinite geometric series in the usual way.
Way 3: Let $p$ be the probability that $5$ comes before $7$. The event "$5$ happens before $7$" can occur in one of two ways: (i) We get a $5$ on the first toss or (ii) we get a $N$ on the first toss, but in the subsequent tossing, $5$ comes before $7$.
Let $p$ be the probability that $5$ happens before $7$. Then from the above analysis
$$p=\frac{4}{36} +\left(\frac{26}{36}\right)p$$
Solve this linear equation for $p$. We get $p=2/5$.
Problem 2: The event "the letters are the same" can happen in one of several ways: (i) we get an S from the first word and an S from the second; (ii) we get a T from the first word and a T from the second; (iii) we get an A from the first word and an A from the second; (iv) we get an I from the first word and an I from the second.
What is the probability of (i)? The probability of getting S from STATISTICS is $3/10$. The probability of getting S from ASSISTANT is $3/9$. So the probability of getting S from the first word and S from the second is $(3/10)(3/9)$.
What is the probability of (ii)? The probability of getting T from the first word is $3/10$. The probability of getting T from the second is $2/9$. So the probability we get a T from each is $(3/10)(2/9)$.
Similarly, the probability of (iii) is $(1/10)(2/9)$ and the probability of (iv) is $(2/10)(1/9)$.
Add up. The required probability is
$$(3/10)(3/9)+ (3/10)(2/9)+ (1/10)(2/9)+ (2/10)(1/9)$$
Please check the numbers, I am not particularly good at counting.
Problem 3: The logic is somewhat like the one of the previous problem. We tossed the die, and got one of $1$, $2$, \dots, $6$, each with probability $1/6$.
"All are red" can happen in several ways. Maybe we tossed a $1$, and got a red on the one ball we drew. If we draw one ball, the probability it is red is $6/10$. So the probability we tossed a $1$, therefore drew one ball, and it was red is $(1/6)(6/10)$.
Maybe we tossed a $2$, and drew $2$ red balls. The probability of drawing red then red is $(6/10)(5/9)$. So the probability we tossed a $2$ and then got $2$ red is $(1/6)(6/10)(5/9)$.
Or maybe we tossed a $3$ then drew $3$ red balls. The probability of this is
$(1/6)(6/10)(5/9)(4/8)$.
Continue, and at the end add up the six probabilities you have computed.
Problem 4: Won't do this, a solution to it has been posted. Also, I don't think the problem is well defined, since we don't know by what process the $2$ surviving letters were obtained.
Let stage $k=0$ denote the person who started the chain (say A) sending out the first two letters to two different people. I assume you are asking for the probability that after stage $k=n$, A does not receive a letter.
Further, as per the problem statements, even if a person receives $m>1$ letters in a stage or across stages, all that is taken care of is that each letter results in a pair of letters going out to two separate people, irrespective of whether they were recipients of other letters by the same sender.
In this situation, we worry only about how many such letter pairs are sent across stages $k=1,2, ...n$. This is $L = 2+4+... 2^n = 2(2^n-1)$.
Each time a letter is received, there are $(N-1)(N-2)$ ways to select two separate people to send a letter pair to. If A is not to be one of them, this choice reduces to $(N-2)(N-3)$. Hence the probability for A not receiving a letter at this time is
$p = \dfrac{(N-2)(N-3)}{(N-1)(N-2)} = \dfrac{N-3}{N-1}$.
As this event occurs independently $L$ times, the total probability is $p^L = \left( \dfrac{N-3}{N-1}\right)^{2(2^n-1)}$.
Best Answer
There is not enough information given to answer the question, since we don't know the prior probabilities of letters arriving from London or from Clifton. A reasonable assumption might be that letters are equally likely to come from any of the people living in those two places. London has a population of roughly $10$ million; the suburb Clifton of Bristol (assuming that that's the Clifton that is meant) has a population of roughly $10{,}000$. Thus the prior probabilities are roughly $0.999$ for arriving from London and $0.001$ for arriving from Clifton.
Regarding the letters on the postmark, again not enough information is given, since we don't know how these letters came to be visible and how likely they are to be from various parts of the words. A reasonable assumption is that a pair of consecutive letters was uniformly randomly selected from all such pairs. Then the probability of a letter from London showing "ON" would be $2/5$, and the probability of a letter from Clifton showing "ON" would be $1/6$.
Under these two assumptions, we can calculate the a posteriori probability of the letter having arrived from London as
$$ P(\text{London}\mid\text{ON})=\frac{P(\text{London}\cap\text{ON})}{P(\text{ON})}=\frac{0.999\cdot\frac25}{0.999\cdot\frac25+0.001\cdot\frac16}\approx0.9996\;. $$
Thus we are now even more certain that the letter must have come from London than before.
If you want to solve this very badly posed problem the way it may have been intended, you may want to use the highly unrealistic prior probabilities $1/2$ instead.