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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$i \in \mathbb{Z}$. 1 Explicitly we have$H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) = \begin{cases} \mathbb{Z}_2 \oplus \mathbb{Z}/2\mathbb{Z} & \text{if } p=2 \newline \mathbb{Z}_p & \text{if } p > 2.\end{cases}$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}\$.