Elliptic curve E and Galois representation Assume that an elliptic curve $E$ over $\Bbb Q$ has a reducible mod $p$ representation. Then 
Q: Why is the semi-simplification of $E[p]$ the direct sum of ${\Bbb Z}/p{\Bbb Z}$ and $\mu_p$?
Next 
Q: How can we find a curve $E'$ isogenous to $E$ such that $E'$ has a point of order $p$ as well as two independent rational point of order $2$.
This occurs in checking that for prime $p > 5$, the mod $p$ representation
$$
\rho_{E_{abc},p} \colon {\mathrm{Gal}}(\overline{{\Bbb Q}}/{\Bbb Q}) \to 
{\mathrm{GL}}_2({\Bbb F}_p)
$$
is irreducible, where $E_{abc}$ is Frey-Hellegouarch curve 
$E_{abc} \colon y^2 = x(x - a^p)(x + b^p)$ associated to triple $a,b,c \in {\Bbb Z}$ such that $a \equiv 3 (4), b \equiv 0 (2)$, and $abc \not=0$. 
To show this, assume to the contrary that $\rho_{E_{abc},p}$ is reducible, then the semi-simplification of $E_{abc}[p]$ should be the direct sum of ${\Bbb Z}/p{\Bbb Z}$ and $\mu_p$. Then, we have $E'$ isogeneous to $E_{abc}$ having the property that $E'$ has a rational point of order $p$ as well as two independent rational point of order $2$, which will be shattered by Mazur's theorem. Thus we see $\rho_{E_{abc},p}$ is irreducible as desired. 
I cannot understand this reasoning, which caused my questions. Pierre 
 A: For your first question, if your elliptic curve is semistable, then what you write is true (and fortunately for your application Frey curves are semistable).  To see this, if E[p] is reducible, we have an exact sequence:
$$
0 \to A \to E[p] \to B \to 0
$$
where $A$ and $B$ are just isomorphic to ${\bf F}_p$, but stable under the action of $G_{\bf Q}$.  Let $\chi_A$ and $\chi_B$ be the corresponding ${\bf F}_p^\times$-valued characters giving the action on $A$ and $B$ respectively.  Since the determinant of the Galois representation on $E[p]$ is $\omega$, the mod $p$ cyclotomic character, we have $\chi_A \cdot \chi_B = \omega$.  We can then write $\chi_A = \omega \chi$ and $\chi_B = \chi^{-1}$.  
Now I claim that the character $\chi$ is unramified away from $p$ in the semistable case.  Indeed, for $q \neq p$, if $q$ doesn't divide the conductor of $E$, then the representation is unramified at $q$ and we are fine.  If $q$ divides the conductor of $E$, since $E$ is semistable, $E$ is a Tate curve at $q$.  In particular, the Galois representation $E[p]$ locally at $q$ has an unramified quotient.  Thus, either $\omega \chi$ or $\chi$ is unramified at $q$ which implies $\chi$ is unramified at $q$ (since $\omega$ is unramified at $q$).
But we can also apply this argument at $q = p$ as $E$ is still a Tate curve at $p$.  In this case, we see that either $\chi$ or $\omega \chi$ is unramified at $p$.  (This argument is essentially Lemma 6 in Serre's article: Propriétés galoisiennes des points d'ordre fini des courbes elliptiques.)
In the first case, $\chi$ is unramified everywhere and thus trivial.  In the second case, $\omega \chi$ is unramified everywhere and $\chi = \omega^{-1}$.  Thus $( \chi_A, \chi_B)$ equals either $(1, \omega)$ or $(\omega,1)$ and $E[p]$ is an extension of $\mu_p$ by ${\bf Z}/p{\bf Z}$ or vice-versa.
On to your second question: the 2-torsion part is easy to handle in the case of Frey curves.  Their explicit Weierstrass equation visibly gives two independent points of order 2.  For the point of order $p$, if either $E[p]$ is a split extension or if ${\bf Z}/p{\bf Z}$ is a sub of $E[p]$ then we are immediately done without applying an isogeny.  If $E[p]$ is a non-split extension with $\mu_p$ as the sub, then simply consider the isogenous curve $E' = E/\mu_p$.  Then $E'[p]$ contains ${\bf Z}/p{\bf Z}$ as sub and we are done.
A: Both questions have incorrect expectations. This has already been noted for the second question by S. Carnahan in the comments. For the first question, take any elliptic curve for which the semisimplification of $E[p]$ is $\mathbb{Z}/p\mathbb{Z} \oplus \mu_p$ and twist $E$ by a quadratic character $\chi\colon \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \rightarrow \{\pm 1\}$ to arrive at a curve $A$ over $\mathbb{Q}$ with $A[p]$ reducible, yet $A[p]^{ss} \cong \chi \oplus \mu_p \chi$.
