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.