In the paper "Fermat's Last Theorem" by H. Darmon, F. Diamond, R. Taylor (2007), theorem 2.15 says that for $l>3$ the Galois representation $\rho_{E.l}$ is irreducible.
My question in general: Is the Galois representation always not irreducible at $l =2$ or $3$, for all elliptic curves?
If it is not, let $E :y^2=x^3+ax+b$ where $ab \in \mathbb Q$ (and we know the value of $ab$, not all elliptic curve). What are the steps to prove that the Galois representation for this specific curve is irreducible at $l=2$?
For an elliptic curve over $\Bbb{Q}$ or any field of characteristic $0$
$$x^3+ax+b = (x-e_1)(x-e_2)(x-e_3)$$ then since $-(x,y) = (x,-y)$ we see $(\infty,\infty),(e_1,0),(e_2,0),(e_3,0)$ is in the 2-torsion and it is the whole of it so $\rho_{E,2}$ (the $GL_2(\Bbb{F}_2)$ action of $G_\Bbb{Q}$ on $E[2] \cong \Bbb{Z/2Z}\times \Bbb{Z/2Z}$) is the Galois group of $x^3+ax+b$'s splitting field. There are 4 cases : $x^3+ax+b$ splits completely over $\Bbb{Q}$, or it is a quadratic times a linear, or it is irreducible with $-4a^3-27b^2$ not a square, or irreducible with $-4a^3-27b^2$ a square.
In the two latter cases the Galois group acts transitively on the points of order $2$ , thus on the 3 subspaces of $\Bbb{F}_2^2$, so $\rho_{E,2}$ is irreducible over $\Bbb{F}_2$. When the discriminant is a square the splitting field is a degree 3 cyclic extension $Gal(K/\Bbb{Q}) = \{1,\sigma,\sigma^2\}$ and (identifying $e_j$ with the point $(e_j,0) \in E[2]$) $$E[2] = e_1\Bbb{F}_2+e_2\Bbb{F}_2,e_1+e_2=e_3,\rho_{E,2}(\sigma)= \pmatrix{0 & 1 \\ 1 & 1}$$ and the characteristic polynomial of this matrix is $x(x+1)+1=(x-\zeta_3)(x-\zeta_3^2)$ which is separable thus it is conjugate to a diagonal matrix over $\Bbb{F}_2(\zeta_3)$ and $\rho_{E,2}$ is reducible over $\Bbb{F}_2(\zeta_3)$.
When $-4a^3-27b^2$ is not a square then $Gal(K/\Bbb{Q})\cong S_3$, it contains an element of order 2 sending $e_1$ to itself and permuting $e_2$ and $e_3=e_1+e_2$ so its matrix is $\pmatrix{1 & 0 \\ 1 & 1}$ and the representation is irreducible over $\overline{\Bbb{F}}_2$.
When $x^3+ax+b= (x+cx+d)(x-e_1)$ then the representation factorizes through $Gal(\Bbb{Q}(\sqrt{c^2-4d}/\Bbb{Q})= \{1,\sigma\}$ $$\rho_{E,2}(\sigma) = \pmatrix{1 & 0 \\ 1 & 1}$$ this matrix is not conjugate to any diagonal matrix thus $\rho_{E,2}$ is irreducible over $\overline{\Bbb{F}}_2$.
When $x^3+ax+b$ splits completely then the representation is trivial and reducible.