Chinese Remainder Theorem, Miller-Rabin Primality test, and more...

Question

Good day, I was going over the proof of the Miller-Rabin Primality Test and have a few questions regarding it. THE BOOK IS COMPLEXITY AND CRYPTOGRAPHY: AN INTRODUCTION

Where

$B_t = \{a\in\Bbb Z_n^* |a^{m·2^t} = \pm1\bmod n \}$

and

$t=\max [0≤i≤k−1|\exists a \in \Bbb Z_n^*\ \text{such that } \ a^{m·2^i} =−1\bmod n ]$

My first question concerns the last few lines. "The definition of t implies that...".

If we say that $n=cd$, how do we show that $b^{m·2^t} \neq \pm1 \bmod n$ using the Chinese Remainder Theorem?

My second question

If $a$ is coprime to n and

$b = a \bmod c$ and $b = 1 \bmod d$

Then how do we show that b is coprime to n?

$n> d,c \geq 3$

Answer 1

1) Since $a^{m.2^t} \equiv -1 \bmod n$ this gives $a^{m.2^t} \equiv -1 \bmod c$ and thus $b^{m.2^t} \equiv -1 \bmod c$. Obviously $b^{m.2^t} \equiv 1 \bmod d$. However this pair of results is not consistent with $b^{m.2^t} \in\{-1,1\} \bmod n$ and thus $b\not \in B_t$.

2) Since $a$ is coprime to $n$, it is also coprime to $c$ and thus $b$ is coprime to both $c$ and $d$.