Is such modulus trick possible? If we have a result of $a \pmod {10^{100} + 7}$ or $a \pmod {10^{100} + 1}$ without knowing what $a$ is, is there a way to get a result of $a \bmod 10^{100}$?
 A: The integers $10^{100}$ and $10^{100}+m$ are coprime whenever $m$ is not divisible by either $2$ or $5$. In particular when $m=1$ or $m=7$. Given this the Chinese Remainder Theorem tells us that the pair of congruences
$$
a\equiv x\pmod{10^{100}},\qquad a\equiv y\pmod{10^{100}+m}
$$
has an integer solution $a$ for ALL pairs $x,y$.
This means that knowing the remainder of an integer modulo $10^{100}+m$ gives no information whatsoever about the remainder modulo $10^{100}$. In other words, the answer to your question is 

"No! It can be anything!"

A: No, by CRT (Chinese Remainder) all pairs of remainders are possible. In fact it is not difficult to explicitly solve this  system:  $ $ let  $\,a=6,\,\ m = 10^{100}+1\ $ below.
$\!\begin{align}
{\bf Theorem}\quad & x\equiv c \pmod{m = ab\!-\!1}\\
& x\equiv d \pmod{n = m\!+\!a}\end{align}\!\! \iff x\equiv d + (c\!-\!d)bn \pmod{mn}$
Proof $\ {\rm mod}\ m\!:\ x\equiv d\iff x = d + k n = d + k(m\!+\!a)\, $ for some $\,k\in\Bbb Z.$
${\rm mod}\ m\!:\ c\equiv x\equiv d+k(m\!+\!a)\equiv d+ka\iff \color{#c00}k\equiv (c\!-\!d)/a\equiv \color{#c00}{(c\!-\!d)b}\,$ by $\,ab\equiv 1\,\Rightarrow\,a^{-1}\equiv b$
Therefore $\ x = d + \color{#c00}kn = d + n(\color{#c00}{(c\!-\!d)b}+jm)\equiv d+(c\!-\!d)bn\pmod{mn}$ 
