Explicitly we have $H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) = \begin{cases} \mathbb{Z}_2 mathbb{Z}_p \oplus \mathbb{Z}/2\mathbb{Z} mathbb{Z}/p\mathbb{Z} & \text{if } p=2 p \le 3 \newline \mathbb{Z}_p & \text{if } p > 2.\end{cases}$3.\end{cases}$
This follows from the Remark following Prop. 7.3.10 of Neukirch et. al (same book as in Timo's answer): $$H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) \cong H^1(\mathbb{Q}_p,\mathbb{Q}_p/\mathbb{Z}_p(-1))^\vee \overset{7.3.10}{\cong} (\mathbb{Q}_p/\mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z})^\vee = \mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z}$$ where $\vee$ denotes the Pontryagin dual and $n=w_p^2$ is the maximal $p$-power such the the degree of $\mathbb{Q}_p(\mu_n) \mid \mathbb{Q}_p$ divides $2$.
The same argument can be used to compute $H^1(K,\mathbb{Z}_p(i))$ for all finite extensions $K \mid \mathbb{Q}_p$ and all $i \in \mathbb{Z}$.
