Answer
a-
(i)one-to-one
(ii) not onto
b-
onto
Work Step by Step
$a-\\
g:\mathbb{Z}\rightarrow \mathbb{Z},g(n)=4n-5\\
A\,function\,\,g: X \rightarrow Y\,is\,\,one-to-one\,\Leftrightarrow \\
\forall \,\,x_{1}\,and\,x_{2}\,\,in\,\,X\,\,if\,
g(x_{1}) = g(x_{2})\,\,then\,x_{1} = x_{2}.\\
so\,let\,g(x_{1})=g(x_{2})\\
\therefore 4x_{1}-5=4x_{2}-5\Leftrightarrow 4x_{1}=4x_{2}\Leftrightarrow x_{1}=x_{2}\\
\therefore g\,is\,one-to-one\\
f\,is\,not\,onto\,as\,20\in \mathbb{Z}\,\\
is\,not\,image\,of\,some\,element\,in\,\mathbb{Z}\\
(because\,4n-5=20\Leftrightarrow 4n=25\Leftrightarrow n=\frac{25}{4}\notin \mathbb{Z})\\
b-\\
G:\mathbb{R}\rightarrow \mathbb{R},G(n)=4n-5\\
G: \mathbb{R} \rightarrow \mathbb{R} \,\,is\,\,onto\,\Leftrightarrow \,\\
\forall x\,in\,\mathbb{R} ,\exists y \in \mathbb{R}\,such\,that\, G(x) = y.\\
so\,let\,y\in \mathbb{R}\\
define\,\,x=\frac{y+5}{4}\\
by\,def.\,of\,G(x)\\
G(x)=4(\frac{y+5}{4})-5=y+5-5=y\\
so\,for\,any\,y\in \mathbb{R}\\
y\,is\,image\,of\,some\,element\,in\,\mathbb{R}
$
