The CRT doesn't apply right away because your moduli are not relatively prime. So there my not be a solution. If the last congruence had been $x\equiv 2 \pmod{4}$ then this would contradict the first congruence.
So in cases like this, one has to be observant and note that any solution to $x \equiv 3 \pmod{4}$ also satisfies $x\equiv 1 \pmod{2}$. So your first congruence is superfluous, and you can delete it. In more complicated cases, you can proceed in gimusi's answer. If you hit a contradiction doing that, then there is no solution.
Below I show how to easily find the errors. Recall (read!) that the reason the CRT formula works is because each summand has the sought value for one modulus, and is $\equiv 0\,$ for all others. Thus your last summand $\,s = \color{#0a0}{13}\cdot 2\cdot\color{#c00} 5\,$ should satisfy $\,s\equiv 0 $ mod $3\ \&\ 4$, and have the sought value mod $23$, i.e. $\,s\,$ should be a root of $\,17\:\! s\equiv 9\pmod{\!23}$.
But your $\,s\not\equiv 0 $ mod $3\ \&\ 4$. The CRT formula achieves that by including a $\rm\color{#0a0}{first\ factor}$ of $\,3\cdot 4 = 12$, but your first factor is $\color{#0a0}{13}$. Fixing that typo your summand becomes $\,s = 12\cdot 2\cdot\color{#c00} 5$.
Finally $\,s\,$ must be a root of $17s\equiv 9\pmod{23}\,$ but yours has $17s\equiv 15\not\equiv 9$. The CRT formula achieves that by choosing a root $\,r\,$ then writing $\,s = 12\:\!(12^{-1}\bmod 23)\:\!r\equiv r.\,$ Your 2nd factor $\,12^{-1}\equiv 2\,$ is correct but your $\rm\color{#c00}{3rd\ factor}$ $\,r\equiv \color{#c00}5\,$ is not a root since $17\cdot 5\equiv 17\not\equiv 9$. Let's fix that by calculating a root $\,r\,$ by twiddling to exact quotients
$$\bmod 23\!:\,\ 17r\equiv 9\iff r\equiv \dfrac{9}{17}\equiv\dfrac{9}{-6}\equiv\dfrac{-3}{2}\equiv\dfrac{20}2\equiv 10\qquad\qquad$$
Thus the correct summand for modulus $\,23\,$ is $\,s = 12\cdot 2\cdot 10$.
Notice how a good understanding of the reason that the CRT formula works allowed us to easily troubleshoot the problem. This is true in general - if you understand the idea behind a proof or formula then you can debug an erroneous application of it be going through the proof line-by-line to determine the first place where the proof breaks down in your special case. For more examples of this debugging method see a "proof" that $1 = 0$ and a "proof" that $2 = 1$.
Below I explain from a general viewpoint the method used in sirous's answer.
$\begin{align}\ 17x&\equiv 9\!\!\!\pmod{\!276}\\[.2em]
\iff\qquad \color{#c00}{17}x {-}\color{#0a0}{276}k &= 9,\ \, {\rm note}\ \,\color{#0a0}{276\equiv 4}\!\!\!\pmod{\!\color{#c00}{17}},\,\ \rm so\\[.2em]
\iff\!\:\! \bmod \color{#c00}{17}\!:\ \ \ {-}\color{#0a0}4k&\equiv 9\equiv -8\iff \color{#c00}{k\equiv 2}\\[.3em]
\iff\:\! \qquad\qquad\quad\ \ x\, &=\, \dfrac{9\!+\!276\color{#c00}k}{17} = \dfrac{9\!+\!{276}(\color{#c00}{2\!+\!17j})}{17} \equiv 33\!\!\!\!\pmod{\!276}
\end{align}$
The above method may be viewed a bit more conceptually as computing a value of $\,\color{#c00}k\,$ that makes exact the following quotient $\, x\equiv \dfrac{9}{17}\equiv \dfrac{9+276\color{#c00}k}{17}\pmod{\!276},\,$ cf. inverse reciprocity.
Best Answer
If $x\equiv 2\pmod 7$ then $x = 2+7k_1$ for some $k_1 \in \mathbb{Z}$. So substituting in your second equation you get $2+7k_1\equiv 4 \pmod {11}$, which gives you $k_1 \equiv 5\pmod{11}$, because $7^{-1}\equiv8\pmod{11}$. Hence $k_1 = 5+11k_2 $, for some $k_2 \in \mathbb{Z}$, and therefore $$x = 2+7(5+11k_2)=37+77k_2\equiv37\pmod{77} $$