Prove that $\sum\Lambda (n)\sim \frac{x}{\varphi(q)}$ where the sum is over every $n\leq x$ such that $n\equiv a\pmod q$ 
Prove that $\sum\Lambda (n)\sim \frac{x}{\varphi(q)}$ where the sum is over every $n\leq x$ such that $n\equiv a\pmod q$

Attempts
Let $\chi$ a character of Dirichlet modulo $q$ non trivial. I proved that $$\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n}=\mathcal O(1)$$
I want now prove that $$\sum_{n\leq x}\Lambda (n)\chi(n)=o(x).$$
Using Abel summation, I get
$$\frac{1}{x}\sum_{n\leq x}\Lambda (n)\chi(n)=\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}-\int_1^x\frac{1}{t^2}\sum_{n\leq t}\Lambda (n)\chi(n)dt $$
Q1) How can I get that $$\int_1^x \frac{1}{t^2}\sum_{n\leq t}\Lambda (n)\chi(n)dt=o(x),$$ to conclude ?
I suppose the previous result correct.
$$\sum_{\substack{n\leq x\\ n\equiv a\pmod q}}\Lambda (n)=\sum_{n\leq x}\Lambda (n)\delta_{1\equiv a\pmod q}=\frac{1}{\varphi(q)}\sum_{n\leq x}\Lambda (n)\sum_{\chi\pmod \chi}\chi(n)\overline{\chi(a)}$$
$$=\underbrace{\frac{1}{\varphi(q)}\sum_{n\leq x}\Lambda (n)}_{\sim\frac{x}{\varphi(q)}}+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0}}\overline{\chi(a)}\sum_{n\leq x}\Lambda (n)\chi(n)$$
$$=\frac{x}{\varphi(q)}+o\left(\frac{x}{\varphi(q)}\right)+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}o(x)$$
$$=\frac{x}{\varphi(q)}+o\left(\frac{x}{\varphi(q)}\right)+o\left(\frac{x}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}\right)$$
Q2) How can I conclude ? i.e. that $$o\left(\frac{x}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}\right)=o\left(\frac{x}{\varphi(q)}\right)\ \ ?$$
 A: 
$\sum_{n < x,  n \equiv a\bmod q} \Lambda(n) \sim \frac{x}{\phi(q)}$ is the prime number theorem in arithmetic progressions. 

Assuming you showed $L(s,\chi)$ doesn't vanish at $s=1$,
You'll need to prove that $L(s,\chi)$ doesn't have any zeros on $\Re(s) = 1$. 
For this let $$F_q(s) = \prod_{\chi \bmod q} L(s,\chi)  = \prod_{p \,\nmid\, q,\ p^k \equiv 1 \bmod q} \frac{1}{(1-p^{-sk})^{\phi(q)/k}}$$ which is a Dirichlet series analytic except a simple pole at $s=1$, and whose logarithm 
$$\log F_q(s) = \phi(q) \sum_{n \equiv 1 \bmod q} \frac{\Lambda(n)}{\ln n} n^{-s}$$ has non-negative coefficients.
Thus we can apply the same argument as for $\zeta(s)$, to obtain a contradiction from $F_q(1+iy)=0$. Thus each $L(s,\chi)$ doesn't vanish on $\Re(s) \ge 1$,
and this is all we need. Let
$$G_{a,q}(s) = \frac{1}{\phi(q)}\sum_{\chi \bmod q} \chi(a) \frac{L'(s,\chi)}{L(s,\chi)}= \sum_{n \equiv a \bmod q} \Lambda(n) n^{-s}\\=s \int_1^\infty (\sum_{qm+a < x} \Lambda(qm+a))x^{-s-1}dx$$
By Mellin inversion 
$$\int_1^x (\sum_{qm+a< y} \Lambda(qm+a)-y) dy \qquad\qquad\qquad\qquad\\= \frac{1}{2i\pi} \int_{\sigma(T)-i T}^{\sigma(T)+iT} (\frac{1}{s-1}-G_{a,q}(s))\frac{x^{s+1}}{s(s+1)}ds +\mathcal{O}(x^{1+\sigma(T)} T^{-1+\epsilon})= o(x^2)$$
(we needed also some bounds $G_{a,q}(\sigma(t)+it)= \mathcal{O}((\log t)^k)$ when $\sigma(t) \to 1$ fast enough)
A: 1) You can use Abel with $a_n=\frac{\Lambda (n)\chi(n)}{n}$ and $f(t)=t$ what will gives you $$\sum_{n\leq x}\Lambda(n)\chi(n)=\sum_{n\leq x}\frac{\Lambda (n)\chi(n)n}{n}=x\sum_{n\leq x}\frac{\chi(n)\Lambda (n)}{n}-\int_1^x\sum_{n\leq t}\frac{\Lambda (n)\chi(n)}{n}\mathrm dt.$$
Since $$\lim_{x\to \infty }\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}$$ exist, there is $\ell\geq 0$ s.t. $$\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=\ell+o(1),$$
and thus $$\sum_{n\leq x}\Lambda (n)\chi(n)= \ell x+o(x)-\int_1^x(\ell+o(1))\mathrm dt=o(x).$$
2) You missed the condition $(n,q)=1$. \begin{alignat*}{1}
\sum_{\substack{n\leqslant x \\ n\equiv a \mod (q)}}\Lambda(n) = & \ \frac{1}{\varphi(q)}\sum_{\chi \mod (q)}\overline{\chi(a)}\sum_{n\leqslant x}\Lambda(n)\chi(n) \\ = & \frac{1}{\varphi(q)}\sum_{\substack{n\leqslant x \\ (n,q)=1}}\Lambda(n)+\frac{1}{\phi(q)}\sum_{\chi\neq 1}\sum_{n\leqslant x}\Lambda(n)\chi(n),
\end{alignat*}
the RHS second term of the las equality gives you a $o\left(\frac{x}{\phi(q)}\right)$ by the previous question, and the first term is $$\sum_{\substack{n\leq x\\ (n,q)= 1}}\Lambda (n)=\sum_{n\leq x}\Lambda (n)-\sum_{\substack{n\leq x\\ (n,q)>1}}\Lambda (n)$$
and $$\sum_{\substack{n\leq x\\ (n,q)>1}}\Lambda (n)=\sum_{p\mid q}\log(p)\sum_{p^k\leq x} 1\ll \log(x)\log(q). $$
The claim follow.

EDIT
The existence of $\lim_{x\to \infty }\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}$ comme from the fact that $$-\frac{L'(\chi,s)}{L(chi,s)}=\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n^s}$$
for $\Re(s)>1$ and since $\frac{L'(\chi,s)}{L(\chi,s)}$ is analytic on $\mathbb Re(s)>0$ and that $L(\chi,1)\neq 0$, using Neuwman theorem allow you to conclude on the existence of $\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n}.$
