The continued fraction expansion is related to the Gauss transformation $T:(0,1)\to(0,1)$, defined by
$$ Tx:=\frac{1}{x} \mod 1. $$
(Indeed, if $x=[a_1,a_2,\ldots)$, then $Tx=[a_2,a_3,\ldots)$.)
It is well known that $T$ admits an absolutely continuous invariant probability measure $\mu$, given by
$$ \mu(A):=\frac{1}{\ln 2}\int_A \frac{dx}{1+x}, $$
and that $T$ is ergodic for $\mu$.
Now, given $M\ge 2$, the set $B$ of $x\in(0,1)$ for which both $a_1$ and $a_2$ are stricly larger than $M$ clearly satisfies $\mu(B)>0$. Hence, by ergodicity, for $\mu$-almost every $x$ there exist infinitely many integers $n$ such that $T^{n-1}x\in B$. This exactly means that $\mu(C)=1$, where $C$ is the set of numbers $x\in(0,1)$ for which there exist infinitely many integers $n$ satisfying both $a_n>M$ and $a_{n+1}>M$.
The sets you consider in 1. and 2. are of the form ${\cal I}(i_n, v_n, M)$ where the sequence $(i_n)$ never hits two consecutive integers. In these sets, only the numbers $a_{i_n}$ are allowed to exceed $M$, hence ${\cal I}(i_n, v_n, M)\cap C=\emptyset$. Then these sets ${\cal I}(i_n, v_n, M)$ are included in the complement of $C$ which is $\mu$-negligible, and it follows that they havee zero Lebesgue measure.
Best Answer
Your set has measure zero by theorems of Khinchin. First a theorem of Khinchin shows that for almost all real numbers $x$ (i.e. outside a set of measure zero) one has $$ \lim_{n\to \infty} \frac{\log q_n}{n}= C $$ for a positive constant $C$. Here $p_n/q_n$ are the convergents of $x$. So almost surely, the denominators $q_n$ are exponential in $n$. In fact the constant $C$ here is known to be $\pi^2/(12 \log 2)$ -- the Levy or Khinchin-Levy constant.
Next another Theorem of Khinchin shows that for almost all $x$ one has infinitely many approximations $p/q$ with $$ \Big| x-\frac{p}{q}\Big| \le \frac{1}{q^2\log q}. $$ Here what is used is that $\sum_q 1/(q\log q)$ diverges.
Putting both results together (and since continued fractions give best rational approximations) we see that for almost all $x$ we have infinitely many $n$ such that $$ \Big| x-\frac{p_n}{q_n} \Big | \le \frac{C}{n q_n^2}, $$ for some positive constant $C$. Now recall that for any $x$ we have $|x-p_n/q_n|$ is about size $1/(q_nq_{n+1})$ and $q_{n+1}$ is about size $a_{n+1}q_n$. Therefore it follows from the above inequality that for infinitely many $n$ one has $a_n \ge cn$ for some constant $c>0$.
Edit: I just got the chance to look at Khinchin's book Continued Fractions, and he discusses this question explicitly in Theorem 30 of that book (page 63 in the Dover edition). Let $\phi(n)$ be an arbitrary positive function ($n\in {\Bbb N}$). Then the inequality $a_n \ge \phi(n)$ is satisfied for infinitely many $n$ for almost all real numbers $x$, provided $\sum 1/\phi(n)$ diverges. On the other hand if $\sum 1/\phi(n)$ converges, then this inequality is satisfied for only finitely many $n$ (and almost all $x$).