Computation of two covariances between time series (one with both series at same time $t$, one with one lagged time series)

Question

We have a random walk defined as: $$m_t=m_{t-1}+u_t$$ with $u_t$ representing a noise i.i.d. with expectation 0 and variance $\sigma^2_u$. Then, we define: $$p_t=m_t+q_tc$$ where $q_t=+1$ or $-1$ and $c$ is a constant. Notice also that $q_t$ are assumed to be serially independent and independent form both $m_t$ and $p_t$. We also have that $\epsilon_t=u_t+c(q_t-q_{t-1})-\theta \epsilon_{t-1}$, with $\epsilon_t$ representing a zero-mean white noise process and $\theta$ an arbitrary parameter. Could someone help me to understand, step by step, how to get that: $$\operatorname{Cov}(q_t,\epsilon_t)=c$$ $$\operatorname{Cov}(q_t,\epsilon_{t-k})=0, \forall k\geq 1\text{?}$$ As to $\operatorname{Cov}(q_t,\epsilon_t)$, I have just get to the point that, since $E[q_t]=0$ and $E[\epsilon_t]=0$, ${Cov}(q_t,\epsilon_t)={E}[q_t\epsilon_t] = E[q_tu_t+cq_t^2-cq_tq_{t-1}-\theta q_t\epsilon_{t-1}]=0+cE[q_t^2]-0-\theta E[q_t\epsilon_{t-1}]=c\times1 - \theta E[q_t\epsilon_{t-1}]=c-\theta E[q_t\epsilon_{t-1}]$, and I do not know how to show that $\theta E[q_t\epsilon_{t-1}]$ equals $0$. On the other hand, as to $Cov(q_t,\epsilon_{t-k})$, I do not have any clue on how to show it is zero $\forall k\geq 1$