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 $T^{n-1}x\in B$, hence such that both $a_n>M$ and $a_{n+1}>M$. But this prevents $x$ from being in the sets you define in 1. and 2. (in these sets, we cannot see two successive $a_n$ strictly larger that $M$). It follows that these sets have zero $\mu$-measure, hence zero Lebesgue measure.