The answer to P1 is negative, thanks to the recent work of Maynard on bounded gaps between primes.
What Maynard shows is that given any $d$, there exists a k-tuple $h_1,\dots,h_k$ such that for infinitely many $n$, at least $d+1$ of $n+h_1,\dots,n+h_k$ are prime. In fact, the argument shows that for sufficiently large $x$, the number of $n \in [x,2x]$ such that at least $m$ of $n+h_1,\dots,n+h_k$ are prime, and the rest are almost prime in the sense that they have no prime factor less than $x^\varepsilon$ for some small fixed $\varepsilon>0$, is $\gg \frac{x}{\log^k x}$. (Such strengthenings of Zhang/Maynard type theorems are discussed in this paper of Pintz, and also in this Polymath8b preprint.) A standard upper bound sieve (e.g. Selberg sieve or beta sieve) then shows that after removing about $O( x/\log^{k+1} x)$ of these $n$, one can also ensure that none of the numbers between $2(n+h_1)$ and $2(n+h_k)$ are prime. If we let $p_i$ be the first prime greater than or equal to $n+h_1$, then we have $2(n+h_1) \leq 2p_i \leq 2p_{i+d} \leq 2(n+h_k)$, and so we obtain a counterexample to P1 for any $d$.
Unfortunately we don't get an effective rate on P2 this way due to the reliance on the Bombieri-Vinogradov theorem in the work of Maynard etc. However this can likely be removed by following the ideas mentioned near the end of this paper of Pintz, though I did not attempt this. [EDIT: it looks likely that the quantitative version of Maynard's results in this recent paper of Banks, Freiberg, and Maynard will do the trick, after some small modification.]
I would like to maintain my basic position on the other post: that this forum does not do well with questions that frequently change. Since the last version has hit rather close to home, I will remark on it separately, in the hopes that research on it will be more fruitful.
The last version I mention is at this writing called version 5. Rewriting it a little, it suggests that for any selection of $n$ primes $d_i$ less than $M$, and for any prime $x$ less than $M$, the number of multiples of $x$ less than $M^2$ which are also multiples of one or more of the $d_i$ do not get very frequent among such multiples. Specifically, $(x-2)M_x \lt M_d$, where $M_d$ is (the count of) the numbers less than $M^2$ which are multiples of one or more of the $d_i$, and $M_x$ is (the count of) that subset of $M_d$ which are multiples of $x$.
This version is likely to be true for two reasons: the subset which I will also call $M_d$ (a union of sets $A_i$ in the original formulation) is large enough to guarantee at least $M$ many members in it, and the conjectured density $1/(x-2)$ is weak enough that it is likely to hold. I do not have a proof of this version. I do want to point out a simple example to think of for counterexamples.
Let $d_i$ range over $2,3$, and $5$, and let us count multiples of $7$ starting from $1$.
It gets boring quickly, so I will just note some of the changing ratios: (0,1), (0,9), (1,10) (for the multiple 14 occurring among the first 10 numbers not coprime to 30), (1,14), (2,15),(2,20),(3,21),(3,25),(4,26)! The point is that the members of the union are distributed sporadically enough that the fourth multiple of 7 among positive nontotatives of 30 occurs early enough that the ratio $M_x/M_d$ could get larger than 1/7.
How much larger? In the general case, I don't know exactly. Part of what I know is discussed in
Bound the error in estimating a relative totient function, which is concerned with the excess of the actual number of coprimes in an interval compared with the expected number. (This excess corresponds to a deficit in the expected number of nontotatives, or the union of the poster's $A_i$.) Various of the conjectures in this question have touched on issues related with a deeper investigation of a problem I have been looking at for years. The best results available stem from analytic number theory; it is my hope that examining combinatorial approaches will shed some light on this and related problems, including the twin prime conjecture mentioned by the poster.
I recommend references in the MathOverflow question 88777 linked above for the poster, including the 2008 paper of Codeca and Nair mentioned in Alan Haynes's answer. I also ask for the poster's understanding in trying a different method for his/her inquiries. Using a different forum such as math.stackexchange.com may turn out to be fruitful, but understanding the accessible literature should be a first step. Fortunately, a lot of that literature is contained in the questions themselves, as well as reference to that literature. Your search may be quickly aided by using phrases and names from 88777.
Gerhard "Good Luck On Your Studies" Paseman, 2015.05.08
Best Answer
older answer The first counter-example is for $p=17$. The interval of length $40$ starting at $87890$ only yields three integers with least prime factor greater than $17$: the primes $87911,87917$ and also $87929=23\cdot 3823$. So we can use the inteval of length $38$ starting at $87890$ or at $87891$.
I may be mistaken, but I think that there are not any examples for $p=19$ and $p=23$.
On the other hand, of the 89 consecutive integers from $43559563512434$ to $43559563512522$ inclusive, all but one have a prime factor $41$ or less. The exception is $43559563512481=9393910613\cdot 4637$
There are essentially 6 ways to get a run of length $39$ (or 3 up to reflection): One can start at $87890,177980,182342,328130,332492$ or $422582 \mod 510510$.
Discussion (Since somebody asked) For a prime $p$ call an integer $p$-good if it has a prime factor $p$ or less and $p$-bad otherwise. Let $M(p,b)$ be the longest possible run of consecutive integers which are all $p$-good except for $b$ exceptions (called holes). Note that $M(p,b)$ is odd. To find the values of $M(p,b)$ and all intervals attaining it, it suffices to look for runs in the integers from $0$ to $p\sharp-1$ where $p\sharp$ is the product of the primes up to $p$. That is true if we look cyclically or go a reasonable distance past $p\sharp$. I wondered at the particular question asked by the OP (" Is it the case that $M(p_{n-1},2)<2p_n$ for all $n$?" But now I realize that must be inspired by the fact that the $M(p_{n-1},2)\ge 2p_n-1$ is witnessed by the run from $p_{n-1}\sharp-p_n+1$ to $p_{n-1}\sharp+p_n-1$ (with the two bad numbers being $p_{n-1}\sharp \pm 1$).
For my computations I generated a list of the $p-bad$ numbers up to $p\sharp$. This can be done iteratively. I did this up to $p=23$ using a fairly uninspired Maple program on a unimpressive computer. $p=19$ took 9 seconds and $p=23$ took 250. The lists are centrally symmetric but I did not exploit that. A faster language, better program, faster machine, and storage in a file might allow one to go somewhat further. I then made the list of jumps from one $p$-bad number to the next. The jumps for $p=17$ with their multiplicities were
$[[2, 22274], [4, 22275], [6, 26630], [8, 6812], [10, 7734], [12, 4096]$
$[14, 1406], [16, 432], [18, 376], [20, 24], [22, 78], [24, 20], [26, 2]]$.
I then had Maple look for 3 successive gaps adding to at least $40$. There were 3 up to the middle.
I did not care to try $p=29$ by the same method! I looked for $M(23,3)$ using the same list of gaps in hopes of finding something good enough that I could get a good bound for $M(29,2)$ use the Chinese remainder theorem to shift everything preserving the 23-good members and plug a hole by making a 23-bad member 29-good. I didn't find anything by eyeballing. Similarly for $M(23,4)$ using $p=29,31$ and $M(23,5)$ using $p=29,31,37$. I did not think to look if there was something attaining $M(23,5)$ so that two of the holes were separated by 58 so that they could be plugged using up just one prime. But something with 2 of 6 holes separated by 62 did get noticed.
Note: I might have overlooked things. I did not really look at the two largish gaps around $p\sharp$. Also, The two holes with one prime trick might make a good example out of a run which is 2 or 4 less than $M(23,b)$.