Answer
a-
(i)not one-to-one
(ii) not onto
b-
onto
Work Step by Step
$
a-\\
f:\mathbb{R}\rightarrow \mathbb{R},f(x)=x^2\\
A\,function\,\,F: \mathbb{R} \rightarrow \mathbb{R}\,is\,not\,\,\,one-to-one\,\Leftrightarrow \\
\exists \,\,x_{1}\,and\,x_{2}\,\,in\,\,\mathbb{R}\,\,such\,that\,\,
F(x_{1}) = F(x_{2})\,\,and x_{1} \neq x_{2}.\\
f(3)=f(-3)=9\,\,and\,\,3\neq -3\\
so\,f\,is\,not\,one-to-one\\
F: \mathbb{R} \rightarrow \mathbb{R} is\,\,not\,\,onto\,\Leftrightarrow \,\\
\exists y\,in\,\mathbb{R} such\,that\,\forall x \in \mathbb{R}, F(x) \neq y.\\
notice\,that\,-1 \in \mathbb{R}\,\,is\,not\,image\,of\,some\,element\,in\mathbb{R}\\
(since x^2=-1\Rightarrow x=\sqrt{-1}\notin \mathbb{R})\\
b-\\
K:\mathbb{R}^+\rightarrow \mathbb{R}^+,K(x)=x^2\\
K: \mathbb{R}^+ \rightarrow \mathbb{R}^+ \,\,is\,\,onto\,\Leftrightarrow \,\\
\forall x\,in\,\mathbb{R}^+ ,\exists y \in \mathbb{R}^+\,such\,that\, K(x) = y.\\
so\,let\,y\in \mathbb{R}^+\\
define\,\,x=\sqrt{y}(y\geq 0)\\
by\,def.\,of\,K(x)\\
K(x)=(\sqrt{y})^{2}=y\\
so\,for\,any\,y\in \mathbb{R}\\
y\,is\,image\,of\,some\,element\,in\,\mathbb{R}
$
