Asymptotics for square-free numbers in an arithmetic progression Set 
$$Q(s,\chi)=\sum_{n=1}^{\infty}\frac{\mu(n)^2\chi(n)}{n^s},\quad (s=\sigma+i\tau),$$
where $\chi$ is a character $\mod q$, 
Show that $Q(s,\chi)=L(s,\chi)H(s,\chi)$ where $H(s,\chi)$ is a bounded, holomorphic function for $\sigma\geqslant \sigma_0>\frac{1}{2}$. Show that $L(s,\chi)\ll_{\varepsilon} (Tq)^{1-\sigma+\varepsilon}$ for $0\leqslant \sigma\leqslant 1$, $|\tau|+1 \leqslant T$ and deduce that, for all $\varepsilon>0$ and uniformly for $x\geqslant 2$, $q\geqslant1$, $(a,q)=1$, one has
$$\sum_{\substack{n\leqslant x \\ n \equiv a \pmod q}} \mu(n)^2 =\frac{x}{q} \prod_{p \nmid q}(1-p^{-2}) + O_{\varepsilon}\big(x^{2/3+\varepsilon} q^{1/3}\big). \qquad\quad(1)$$ 
I find this problem in G.Tenenbaum's book in the page 265.
Gérald Tenenbaum. Introduction to Analytic and Probabilistic Number Theory, Cambridge: Cambridge University Press, 1995.
I can get that
By Euler's Product formula 
$$Q(s,\chi)=\prod_{p}\Big(1+\frac{\chi(p)}{p^s}\Big)=\frac{L(s,\chi)}{L(2s,\chi^2)},$$
$$|\mu(n)^2\chi(n)|\leqslant 1, \quad \text{and} \quad \sum_{n=1}^{\infty}\frac{|\mu(n)^2\chi(n)|}{n^\sigma} \leqslant \zeta(\sigma)$$
so $H(s,\chi)=1/L(2s,\chi^2)$ is a bounded, holomorphic function for $\sigma\geqslant \sigma_0>\frac{1}{2}$. 
By Perron's formula
$$\sum_{n\leqslant x}\mu(n)^2\chi(n) =\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{L(s,\chi)}{L(2s,\chi^2)}\frac{x^s}{s}\mathrm{d}s +O\Big(\frac{x^c}{T}\Big).$$
Then how can I do? How to prove $(1)$?
 A: Instead of using Perron's formula, I will use one of the simplest and most versatile tricks in analytic number theory - the fact that the sum over the Mobius function is $1$  when $n=1$, and $0$  otherwise.That is, we will use the fact that: $$\sum_{d|n}\mu(d)=\begin{cases}
1 & \text{if }n=1\\
0 & \text{otherwise}
\end{cases}.$$  If I am not mistaken, this approach results in a significantly better error term than the one quoted in your question.
Since $\mu^{2}(n)$ is the indicator function for the squarefree numbers, it follows from our Möbius identity that $\sum_{d^{2}|n}\mu(d)=\mu^{2}(d).$ Let $$Q(x;q,a)=\sum_{\begin{array}{c}
n\leq x\\
n\equiv a\ (q)
\end{array}}\mu^{2}(n).$$ Inserting this identity and switching the order of summation we have $$Q(x;q,a)=\sum_{\begin{array}{c}
n\leq x\\
n\equiv a\ (q)
\end{array}}\sum_{d^{2}|n}\mu(d)=\sum_{d^{2}\leq x}\mu(d)\sum_{\begin{array}{c}
n\leq x:\ d^{2}|n\\
n\equiv a\ (q)
\end{array}}1.$$ For the inner sum, $$\sum_{\begin{array}{c}
k\leq x/d^{2}\\
k\equiv d^{-2}a\ (q)
\end{array}}1=\frac{x}{qd^{2}}+O(1),$$ when $d$  is relative prime to $q$, and $0$ otherwise, and so $$Q(x;q,a)=\frac{x}{q}\sum_{\begin{array}{c}
d^{2}\leq x\\
(d,q)=1
\end{array}}\frac{\mu(d)}{d^{2}}+O\left(\frac{\phi(q)}{q}x^{1/2}\right).$$ Now $$\sum_{\begin{array}{c}
d^{2}\leq x\\
(d,q)=1
\end{array}}\frac{\mu(d)}{d^{2}}=\prod_{p\nmid q}\left(1-p^{-2}\right)+O\left(\frac{1}{\sqrt{x}}\right)$$ so we obtain the final result:

$$Q(x;q,a)=\frac{x}{q}\prod_{p\nmid q}\left(1-p^{-2}\right)+O\left(x^{1/2}\right) .$$

