Answer
$\\
a-\\
y\in F(A)\,\,or\,\,y\in F(B)\\
b-\\
some\\
c-\\
A\cup B\\
d-\\
F(A\cup B)\\
e-\\
some\\
f-\\
y\in F(A\cup B)$
Work Step by Step
$Let\,X\,and\,Y\,be\,sets,\,let\,A\,and\,B\,be\,any\,subsets\,of\,X\\
and\,\,
let\,F\,\,be\,a\,function\,from\,X\,to\,Y \\
Fill\,in\,the\,blanks\,in\,the\,
following\,proof\,that\,\\F(A) \cup F(B) \subseteq F(A \cup B).\\
Proof:Let\,y\,be\,any\,element\,in\,F(A) \cup F(B) \\
We\,must
\,show\,\,that\,y\,is\,in\,F(A \cup B).\\
By\,definition\,of\,union\,\,\,y\in F(A)\,or\,y\in F(B)\\
Case 1,\,\,y \in F(A):\\ In\,this\,case,\,by\,definition\,of\,F(A)\\
y = F(x) for\,some\,\,x \in A.\\
Since A \subseteq A \cup B,\\ it\,follows\,from\,
the\,definition\,of\,union\,that\,\,x \in A\cup B.\\
Hence, y = F(x) for
some\,\,x \in A \cup B,\,and\,thus,\,by\,definition\,of\,F(A \cup B)\\
y \in F(A\cup B)\\
$
$Case 2, y \in F(B):\\
In\,this\,case,\,by\,definition\,of\,F(B), some\,\,
x \in B.\\ Since\,B\subseteq A \cup B\,it\,follows\,from\,the\,definition\,of\,
union\,that\,\\
y\in F(A\cup B)\\Therefore,\,regardless\,of\,whether\,y \in F(A)\,or\,y \in F(B),\\
we\,have\,that\,y \in F(A \cup B)\,as\,was\,to\,be\,shown$
