In the question, the same message $m$ is encrypted in different ways using textbook RSA. Don't do this in a real application, because it opens to all kinds of attacks (those discussed in the question, and others). Textbook RSA encryption should only be used as a building block to encrypt individually randomized messages, using a process called random padding. See e.g. RSAES-OAEP.
Common message and modulus attack
The example given for the common modulus attack is not RSA, because $e_1$ and $e_2$ are even, thus not coprime with $p-1$ and $q-1$, as required for $e$ in RSA (and unambiguous decryption).
Further, here $e_2=3\,e_1$, thus $c_2={c_1}^3\bmod n$, thus being given $c_2$ can't be of any help (with $n$, $e_1$ and $e_2$ assumed public, it's trivial to find $c_2$ from $c_1$).
We are left to solve $c_1=m^{e_1}\bmod n$, and that's hard when the factorization of $n$ is unknown, unless $m$ is small enough‡ that $m^{e_1}<n$, or close to that. When $e_1$ is a proper RSA exponent for $n$, and $m$ is essentially random in $[0,n)$ as it should be, that's the RSA problem. Here with $e_1=2$, that problem is extracting a modular square root modulo a large composite $n$ of unknown factorization, a problem that's demonstrably as hard as factorization for random $m$.
That's general: when $\gcd(e_1,e_2)=g>1$ in the common modulus attack, the standard "common modulus" attack leaves us with solving $c=m^g\bmod n$ (or $c=m^{g'}\bmod n$ with a $g'$ multiple of $g$ that we can choose, which is not easier), and we are stuck unless $m^g<n$, or close to that. Extracting modular $g^\text{th}$ root is trivial with $g=1$, but for random $n$ is believed intractable for every other $g$ obtained as the GCD of proper RSA exponents, and for $g=2$ as in the question.
There are attacks with "common modulus" in their name that do not require that the GCD of the $e_i$ is $1$, but AFAIK they require largish $\prod e_i$ and small $d_i$, which can't be the case in the question for safe choice of $n$, since the $e_i$ are few and small.
Common message and exponent attack
In the question's final part, the quantity $m^e$ appears in the two mathematical expressions† of the available ciphertexts $c_1$ and $c_2$, before reduction modulo two known $n_1$ and $n_2$ of unknown factorization. Using the Chinese Remainder Theorem allows to compute $m^e\bmod(n_1\,n_2)$, e.g. as $c=\Bigl(\left(n_2^{-1}\bmod n_1\right)\left(c_1-c_2\right)\bmod n_1\Bigr)\,n_2+c_2$. This is an integer $c$ with $0\le c<n_1\,n_2$.
If it happens that $m^e<n_1\,n_2$, then we must have $m^e=c$ which allows attack‡.
It can be seen that when we have $e$ ciphertexts rather than $2$ in the example, $m^e<\prod n_i$ and thus the attack always succeeds. That's Håstad's broadcast attack. The academically correct way to prevent that attack and it's extensions is not to use a large $e$, or use different $e$ for each public key. It's to use random padding at each encryption, as stated in preliminary.
† Notice that the full $m^e$ is not computed in normal use, because modular exponentiation algorithms used to compute $m^e\bmod n$ perform modular reduction along the way, after each modular multiplication or modular squaring.
‡ If $m^e<n$, then $c=m^e\bmod n\,=\,m^e$ without modular reduction and we can find $m$ by computing $m=\sqrt[e]c$ in standard integers $\mathbb Z$. There are extensions that allow somewhat larger $m$, see this old question that I'm attempting to revive.
