It is easy to show that in RSA, when e = 3 there are 4 messages m for which the ciphertext is equal to the plaintext and gcd(m, n) = 1
Well, if $m^3 = m \pmod n$ (and assuming $n$ is a conventional RSA modulus, that is, it is $n = pq$, for $p, q$ distinct odd primes), this is equivalent to both of the below holding simultaneously:
$$m^3 = m \pmod p$$
$$m^3 = m \pmod q$$
If $p, q$ are prime, these are cubic equations in fields; such cubic equations have (at most) 3 solutions. A moment's reflection (or a bit of algebra) yields the solutions $m = 0, 1, -1$ (with the later being equivalent to $p-1, q-1$) - since there are at most 3 solutions, this must be all of them.
Now, $m=0$ (in either case) is inconsistent with $\gcd(m, n)=1$, hence we can discard those solutions; this yields the solutions $m = 1, -1 \pmod p$ and $m = 1, -1 \bmod q$. By the Chinese Remainder Theorem (and the fact that $p, q$ are relatively prime), all four possible combinations correspond to a single $m$ in the range $(0, n-1)$.
The combination $m = 1 \pmod p$ and $m = 1 \pmod q$ yields the value $m = 1$; similarly the combination $m = -1 \pmod p$ and $m = -1 \pmod q$ yields the value $m = n-1$ (the quote gives this as $-1$, however that's not in the range, and modular cubing will never return the value -1); these are the two trivial solutions.
The other two combinations, both $m = 1 \bmod p$ and $m = -1 \pmod q$, and $m = -1 \bmod p$ and $m = 1 \pmod q$ are the nontrivial solutions.
This logic shows there are no other possibilities.
Also, how to find the other 2 messages when there is no clue about n,p,q?
Even if you were given the value of $n$, knowing one of the two nontrivial values immediately leads to a factorization of $n$, for example, by computing $\gcd(m-1, n)$, hence there is no easy way (without knowing the factorizatoin apriori).
