causal inference exercise - "covariate-specific effect" 
This graph and questions come from: CAUSAL INFERENCE IN STATISTICS A Primer - Pearl Glymour and Jewell (2016).
We are interested in the effect of $X$ on $Y$. In order to identify it we looking for the sets compliant with the backdoor criterion, for example $[A Z]$ and $[A Z C]$ seems me among them. For simplicity we assume that the SCM is linear and the path coefficients are named: $\beta_{y,w}$, $\beta_{y,z}$, …, $\beta_{z,c}$, … and so on.
So the causal effect of $X$ on $Y$ is: $\beta_{y,w} \beta_{w,x}$
I ask you If my solutions for the study question above (a bit extended/modified by me) are correct.
point (a)
We looking for $P(Y=y|do(X=x),C=c)$
I read that, given $(X,Y)$, we need a set of control that include $C$ and deal with backdoor criterion. So $[A Z C]$ is a compliant set. The expanded expression is given in the book but I want to understand if, given the linear simplification, the regression like:
$Y = \theta_1 X + \theta_2 C + \theta_3 A + \theta_4 Z + r$
Is what we need, and in particular if $\theta_1 + \theta_2$ represent the c-specific effect. In term of path coefficients we have $\beta_{y,w} \beta_{w,x} + \beta_{y,d} \beta_{d,c} $.  It is correct?
Point (b)
$[Z A B C D]$ seems me a correct set and a regression similar to the the previous is what we need and the z-specifc effect is: $\beta_{y,w} \beta_{w,x} + \beta_{y,z}$. It is correct?
Point (c)
Let me say that, for example, $\beta_{y,w} \beta_{w,x} = 3,6$ and $\beta_{y,z}=0,9$
so $E[Y|do(x),z] = 3,6 x + 0,9 z$
if $z = (1,2)$ we have $x = 0$ , se $z = (3,4,5)$ we have $x = 1$
if $z = 1$
$E[Y|do(x),z] = 3,6*0 + 0,9*1 = 0,9$
if $z = 2$
$E[Y|do(x),z] = 3,6*0 + 0,9*2 = 1,8$
if $z = 3$
$E[Y|do(x),z] = 3,6*1 + 0,9*3 = 6,4$
if $z = 4$
$E[Y|do(x),z] = 3,6*1 + 0,9*4 = 7,2$
if $z = 5$
$E[Y|do(x),z] = 3,6*1 + 0,9*5 = 8,1$
It is correct?
If I'm wrong what are the correct solutions ?
EDIT: In order to clarify my doubts I add something. I’m sure that in linear models the quantity involved in the equation below (letters have no relations with the example above):

can be translated in regression term as follow. I’have to perform the regression:
$Y=\theta_1 X + \theta_2 W + r$
and $\theta_1$ give us the total effect of $X$ on $Y$. Therefore what I looked for.
Now the z-specific effect of $X$ on $Y$ is representable as: 
But in linear regression terms what regression I have to perform? What coefficient I'm interested in?
In a bit different case, the w-specific effect of $X$ on $Y$ is also representable as:

But in linear regression terms what regression I have to perform? What coefficient I'm interested in?
 A: $\newcommand{\op}[1]{\operatorname{#1}}
\newcommand{\doop}{\op{do}}$
Here's my answer.
a. To get the $c$-specific effect of $X$ and $Y,$ which is $P(Y=y|\doop(X=x),C=c),$ we analyze
as follows. The set $S$ in Rule 2 is actually the $\{Z,C\}$ nodes, minimally. Hence, we have that
$$P(Y=y|\doop(X=x),C=c)=\sum_z P(Y=y|X=x,Z=z,C=c)\,P(Z=z|C=c).$$
b. We would have to measure $X,Y,Z,$ and one of $A,B,C,$ or $D.$ I'll pick $A$, so that the
expression becomes
$$P(Y=y|\doop(X=x), Z=z)=\sum_aP(Y=y|X=x,A=a,Z=z)\,P(A=a|Z=z).$$
c. We have that
$$
X=
\begin{cases}
0,&Z=1,2\\
1,&Z=3,4,5.
\end{cases}
$$
Here $Z\in\{1,2,3,4,5\}.$ Now the desired quantity is $E(Y)$ under the $Z$ strategy. That is,
we want
\begin{align*}
E(Y)
&=\sum_y\left[y P(Y=y|\doop(X=g(Z)))\right]\\
&=\sum_y\left[y \sum_zP(Y=y|\doop(X=x),Z=z)|_{x=g(z)}\,P(Z=z)\right]\\
&=\sum_y\left[y \sum_z\left[\sum_aP(Y=y|X=x,A=a,Z=z)\,P(A=a|Z=z)\right]_{x=g(z)}\,P(Z=z)\right]\\
&=\sum_{a,y}\sum_z\left[y\,P(Y=y|X=g(z),A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\\
&=\sum_{a,y}\Bigg\{
\sum_{z=1}^2\left[y\,P(Y=y|X=0,A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\\
&\qquad+\sum_{z=3}^5\left[y\,P(Y=y|X=1,A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\Bigg\}.
\end{align*}
That's about as far as we can get without knowing the probability distributions more exactly.
