Let us begin with the case $A=B$. Then we have $X = A$ and $ Y= C \cup D$. Notice the long exact sequence for the triple $(A,C,C \cap D)$ is $$ \cdots \rightarrow H_n(C,C \cap D)\rightarrow H_n(A,C \cap D)\rightarrow H_n(A,C) \rightarrow H_{n-1}(C,C\cap D) \rightarrow \cdots $$
Further the long exact sequence for the triple $(A,C \cup D, D)$ is
$$ \cdots \rightarrow H_n(C \cup D, D)\rightarrow H_n(A, D)\rightarrow H_n(A,C\cup D) \rightarrow H_{n-1}(C\cup D, D) \rightarrow \cdots $$
The pivotal observation here is that the quotient spaces $C / (C \cap D) \simeq (C \cup D)/D$ yeilding $H_n(C, C \cap D) \approx H_n (C \cup D, D)$. Thus we have the diagram 
Applying the result of problem 2.2.38 we have the relative Mayer-Vietoris sequence for the pair $(A, C \cup D)$:
$$ \cdots \rightarrow H_n(A, C \cap D) \rightarrow H_n(A,C) \oplus H_n(A,D) \rightarrow H_n(A,C \cup D) \rightarrow H_{n-1}(A,C \cap D) \rightarrow \cdots$$
For the case $C=D$ we must be careful as the relative Mayer-Veitoris is defined for the pair $(X,Y)=(A \cup B,C \cup D)$ where $C \subseteq A$ and $D \subseteq B$. Thus we assume $C \subseteq A \cap B$ and make the following observations of triples to determine the Mayer Veitoris sequence for $(A \cup B,C)$. The long exact sequence for the triple $(A,A\cap B,C)$:
$$ \cdots \rightarrow H_n(A\cap B,C) \rightarrow H_n(A,C) \rightarrow H_n(A,A \cap B) \rightarrow H_{n-1}(A \cap B,C) \rightarrow \cdots$$ And for the triple $(A \cup B,B,C)$:
$$ \cdots \rightarrow H_n(B,C) \rightarrow H_n(A\cup B,C) \rightarrow H_n(A \cup B, B) \rightarrow H_{n-1}(B,C) \rightarrow \cdots$$
Again we deduce $H_n(A \cup B,B) \approx H_n(A,A \cap B)$. Thus we have the diagram:

Applying the result of 2.2.38 we have the relative Mayer-Veitoris sequence for $(A \cup B,C)$:
$$ \cdots \rightarrow H_n(A\cap B, C) \rightarrow H_n(A,C) \oplus H_n(B,C) \rightarrow H_n(A\cup B,C) \rightarrow H_{n-1}(A\cap B,C) \rightarrow \cdots$$