= Solution
In the second <causal directed acyclic graph>, $Y(a_1,a_2)$ depends on $U_2$ but not on $U_1$, whereas $A_1$ depends on $U_1$ and the two latent roots are independent. Hence
$$
Y(a_1,a_2)\mathrel\perp A_1.
$$
Although conditioning on $X$ conveys information about $U_2$, the assignment $A_2$ uses only $(A_1,X)$ and fresh randomization, so
$$
Y(a_1,a_2)\mathrel\perp A_2\mid A_1,X.
$$
These are the two <sequential exchangeability> conditions. Using them successively, together with <consistency of potential outcomes>, gives
$$
\begin{aligned}
\mathbb E[Y(a_1,a_2)]
&=\sum_x\mathbb E[Y(a_1,a_2)\mid A_1=a_1,X=x]
\mathbb P(X=x\mid A_1=a_1)\\
&=\sum_x\mathbb E[Y\mid A_1=a_1,X=x,A_2=a_2]
\mathbb P(X=x\mid A_1=a_1),
\end{aligned}
$$
which is the formula from part a.
Adding $X\to Y$ invalidates the argument in general. The latent variable $U_1$ confounds $A_1$ and $X$, so $\mathbb P(X\mid A_1=a_1)$ need not equal the distribution of $X(a_1)$. When $X$ directly affects $Y$, that discrepancy no longer cancels after summing over $x$. The same observed distribution can then correspond to different intervention means, so the displayed formula need not identify the effect.
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