DPA Attack on HMAC SHA-256
I will keep the same notations as in the paper mentioned above.

First, your substitution for $H^{(0)}$ is correct but you could stop to
\begin{align*}
T_1^{(1)} = H^{(0)}\boxplus \Sigma_1(E^{(0)}) \boxplus Ch(E^{(0)}, F^{(0)}, G^{(0)}) \boxplus K_1 \boxplus W_1
\end{align*}
since all values except $H^{(0)}$ are known from previous DPAs.

Regarding $C^{(0)}$, it is a little less straightforward. We have
\begin{align*}
T_2^{(1)} &amp;= \Sigma_0(A^{(0)}) \boxplus Maj(A^{(0)}, B^{(0)}, C^{(0)})\\
&amp;= \Sigma_0(A^{(0)}) \boxplus \big[(A^{(0)} \wedge B^{(0)}) \oplus (A^{(0)} \wedge C^{(0)}) \oplus(B^{(0)} \wedge C^{(0)})\big]
\end{align*}
which results in
\begin{align*}
\big[T_2^{(1)} \boxminus \Sigma_0(A^{(0)})\big] \oplus (A^{(0)} \wedge B^{(0)})&amp;=  (A^{(0)} \wedge C^{(0)}) \oplus(B^{(0)} \wedge C^{(0)}) \,.
\end{align*}
Unfortunately, as bitwise AND operator is not invertible, we don't get $C^{(0)}$ directly. Still, $C^{(0)}$ is the only unknown variable making a bruteforce attack on this 32-bit variable easy in practice.