Here are some hints:

For 5.22: Since $a \equiv b \pmod{m}$, we have that $m \mid b -a$. We also have that $d\mid a$ and $d\mid b$, so $d \mid b -a$. Thus, there exist integers $k$ and $l$ such that $mk = b -a$ and also $dl = b - a$. Another thing that might help is the fact that we can write $d = \gcd(d,m)\cdot f$, where $\gcd(f,m) = 1$. The goal is to show that $\dfrac{a}{d} \equiv \dfrac{b}{d} \pmod{\dfrac{m}{\gcd(d,m)}}$, so we want to show that $\dfrac{m}{\gcd(d,m)} \mid \dfrac{b - a}{d}$, that is, we want to be able to write $\dfrac{b - a}{d} = \dfrac{m}{\gcd(d,m)} \cdot K$, where $K$ is some integer.

For 5.27: Try using $\pmod{8}$.