$a \equiv b \mod n$ (notice the equivalence sign, $\equiv$ has three bars, not $2$) mean that $a$ and $b$ both have the same remainder when divided by $n$.
So for example, yes $5 \equiv 1 \mod 2$ but also $5 \equiv 7 \mod 2$.
A few things to notice:
1) $a \equiv b \mod n \iff \frac {a-b}n \in \mathbb Z \iff a = b + kn$ for some integer $k$. And $a \equiv 0 \mod n\iff n|a$.
2) For any natural number $n$, we can divide the integers, $\mathbb Z$, into exactly $n$ different subsets, or "classes". Class $[0] = \{...., -2n, -n, 0, n, 2n, 3n,...\}=\{kn + 0|k\in \mathbb Z\}$; these are all the integers that have remainder $0$ when divided by $n$. Class $[1] = \{...,-2n, -n+1, 1, n+ 1, 2n+1,...\}=\{kn + 1|k\in \mathbb Z\}$; these are all the integers that have remainder $1$ when divided by $n$. And so on. Class $[i] = \{...,-2n, -n+i, i, n+ i, 2n+i,...\}=\{kn + i|k\in \mathbb Z\}$. There are exactly $n$ of these classes. $a \equiv b \mod n$ means that both $a$ and $b$ belong to the same one of these classes.
The classes in 2) are called equivalence classes because when doing basic arithmetic in regards to remainders when divided by $n$, if two numbers have the same remainder they are ... well... equivalent that is.
Proposition: If $a \equiv b \mod n$ then $a \pm k \equiv b \pm k$ and $k*a \equiv k*b \mod n$ and $a^k \equiv b^k \mod n$. Also if $c \equiv d \mod n$ then $a+c \equiv b+d \mod n$ and $ac \equiv bd\mod n$.
Ex: $5 \equiv 12 \mod 7$ so $5k \equiv 12k \mod 7$ because $12k = 7k + 5k \equiv 5k \mod 7$.
To solve an equation such as $3x \equiv 6 \mod 9$ can have multiple solutions. $x$ can be $2$ because $3*2 = 6\equiv 6 \mod 9$. But $x$ can also be $5$ because $3*5 = 15 \equiv 6 \mod 9$. $x$ could also be $8$ becase $3*8 = 24\equiv 6 \mod 9$. The next solution is $x = 11$ as $3*11=33\equiv 6 \mod 9$. But because $2 \equiv 11 \mod 9$, $2$ and $11$ are considered to be the same solution because $2$ and $11$ are equivalent (modulo $9$).
So it turns out there are three classes of solutions: $x \equiv 2 \mod 9$, or $x\equiv 5 \mod 9$, or $x \equiv 8 \mod 9$.
So the Chinese Remainder Theorem states:
If you have a system of equations:
$x \equiv a_1 \mod n_1$
$x \equiv a_2 \mod n_2$
....
And all the $n_i$ are relatively prime (have no factors in common) then
$x \equiv b \mod n_1n_2n_3....n_m$ has one specific modulo class mod $n_1n_2n_3...n_m$.
Example:
If $x \equiv 1 \mod 2$
$x \equiv 1 \mod 3$
$x \equiv 0 \mod 5$ then
$x \equiv ???? \mod 30$ has only one solution (from the classes $[0]$ to $[29]$). That solution is $x \equiv 25 \mod 30$.
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So your question:
$n = 3a + 2 \iff n \equiv 2 \mod 3$.
$n = 7b + 5 \iff n \equiv 5 \mod 7$
$n = 8c + 9 \iff n \equiv 9 \equiv 1 \mod 8$.
As $3,7, 8$ are relatively prime, there is one solution modulo $3*7*8= 168$.
$n\equiv 3a + 2 \equiv 7b + 5 \mod 21$ has one possible value (modulo $21$)
$3a + 2$ may be $2,5,8,...., 20$. And $7b + 5$ may be $5,12, 19$. The only option in common is $n \equiv 3a +2 \equiv 7b + 5\equiv 5 \mod 21$. So $n = 21d + 5$ for some $d$.
Now we also have $n \equiv 8(c-1) +1 \mod 21*8$ and $n \equiv 21d + 5\mod 21*8$.
So $8(c-1)+1$ may be $1, 9,17,.....$ and $21d+5$ may be $5, 26, 47, 68, 89, 110, 131, 152$. The only one in common is $89$.
So $n \equiv 3a + 2 \equiv 7b + 5 \equiv 8(c-1) + 1 \equiv 89 \mod 168$
And $a = \frac {87}3; b = \frac {84}7; c = \frac {80}8$.
