Let $(X,A)$ be a finite CW-pair $m=p^r$ for some prime $p$. Unspecified coefficient is in $\mathbb{Z}$. From the universal coefficient theorem, We know that
$H^1(A;\mathbb{Z}_m)=\textrm{Hom} (H_1(A),\mathbb{Z}_m)$ ---(1) and
$H^2(X,A;\mathbb{Z}_m)=\textrm{Hom}(H_2(X,A);\mathbb{Z}_m)\bigoplus \textrm{Ext}(H_1(X,A),\mathbb{Z}_m)$. ---(2)
From the long exact sequence of pair, we have a coboundary map $\delta\colon H^1(A;\mathbb{Z}_m)\to H^2(X,A;\mathbb{Z}_m)$.
I know that $\pi_1\circ \delta \colon H^1(A;\mathbb{Z}_m)\to \textrm{Hom}(H_2(X,A),\mathbb{Z}_m)$ ($\pi_1$ is a 1st factor projection from (2)) is same as the composition $H^1(A;\mathbb{Z}_m)\cong \textrm{Hom}(H_1(A);\mathbb{Z}_m)\to \textrm{Hom}(H_2(X,A);\mathbb{Z}_m)$ (the first map is an isomorphism from (1) and the second map is obtained by taking $\textrm{Hom}(-,\mathbb{Z}_m)$ to $\partial\colon H_2(X,A)\to H_1(A)$.)
Let's think $\pi_2\circ \delta \colon H^2(A;\mathbb{Z}_{m})\to Ext(H_1(X,A);\mathbb{Z}_m)$, where $m=p^r$ as before.
Question : Is it true that if $r$ is large, then $\pi_2\circ\delta$ is trivial map or is it meaningless question because of unnaturality of splitting of (2)?