Answer
a-
f is not one-to-one
f is onto
b-
f is not one-to-one
f is not onto
Work Step by Step
$a-\\
A\,function\,\,F: X \rightarrow Y\,is\,not\,\,\,one-to-one\,\Leftrightarrow \\
\exists \,\,x_{1}\,and\,x_{2}\,\,in\,\,X\,\,such\,that\,\,
F(x_{1}) = F(x_{2})\,\,and x_{1} \neq x_{2}.\\
f\,\,is\,not\,one-to-one \\
as\,f(c)=f(d)=e \,\,\,but\,\,c\neq d\\
f: X \rightarrow Y is\,\,onto\,\Leftrightarrow \,\\
\forall y\,in\,Y ,\exists x \in X\,such\,that\, f(x) = y.\\
so\,f\,is\,onto\,as\,\,\\
e,f\,and\,g(elements\,of\,Y)is\,\,the\,image\,of\,some\,element\,in\,X \\
$
$b-\\
A\,function\,\,G: X \rightarrow Y\,is\,not\,\,\,one-to-one\,\Leftrightarrow \\
\exists \,\,x_{1}\,and\,x_{2}\,\,in\,\,X\,\,such\,that\,\,
G(x_{1}) = G(x_{2})\,\,and x_{1} \neq x_{2}.\\
G\,\,is\,not\,one-to-one \\
as\,G(a)=G(b)=f \,\,\,but\,\,a\neq b\\
G\,is\,not\,one\,\,to\,\,one.\\
G: X \rightarrow Y is\,\,not\,\,onto\,\Leftrightarrow \,\\
\exists y\,in\,Y such\,that\,\forall x \in X, G(x) \neq y.\\
we\,notice\,that\,g\in Y \,\,is\,not\,image\,of\,some\,element\,in\,X
$
