The answer to Q1 is yes, the order of $a$ might be smaller than the gcd: Let $G=\langle x,y\mid x^8=y^2=1,x^y=x^3\rangle$ be the semidihedral group of order $16$. Let $a=[x]\in G_{\text{ab}}=C_2\times C_2$. Clearly $\lvert a\rvert=2$. Let $X$ be the set of subgroups $H$ with $V_H(a)=0$. I claim that $\gcd_{H\in X}([G:H])\geq 4$: (Clearly $1\in X$ so the gcd is welldefined). One just have to see that $X$ does not contain any of the $3$ maximal subgroups $H_1=\langle x\rangle$, $H_2=\langle x^2,y\rangle$ and $H_3=\langle x^2,xy\rangle$ of $G$. If I did the transfer computation correctly one has $V_1(a)=[x^4]\neq 0$, $V_2(a)=[x^2]\neq 0$, $V_3(a)=[x^2]\neq 0$ proving the claim. (One can actually skip $H_3$ here since there is an automorphism of $G$ which swaps $H_2$ and $H_3$.)

**Update**
The answer to Q2 is yes if $G$ is finite (I leave the profinite case to the interested reader....): Let $a\in G^{\text{ab}}$ and $X$ be the set of subgroups of $G$ such that $V_H(a)=0$. Note that $1\in X$, so $\lvert G\rvert a=0$. Assume that $p$ is a prime which does not divide $\text{ord}(a)$, and let $S$ be a Sylow $p$-subgroup of $G$. Then $p$ does not divide $\text{ord}(V_S(a))$. On the other hand $V_S(a)\in S^{\text{ab}}$ has $p$-power order so $V_S(a)=0$. Hence $S\in X$ so $p$ does not divide $\gcd_{H\in X} [G:H]$.