Answer
a-L is not one-to-one
b-L is onto
Work Step by Step
$l:S\rightarrow \mathbb{Z}^{nonneg} (S\,is\,the\,set\,all\,strings\,of\,0\,s\,and\,1\,s)\\
l(s) =\,the\,length\,of\,\,s,\,\,for\,\,all\,\,strings\,s\,in\,S.\\
a-\\
A\,function\,\,l: S \rightarrow \mathbb{Z}^{nonneg}\,is\,not\,\,\,one-to-one\,\Leftrightarrow \\
\exists \,\,x_{1}\,and\,x_{2}\,\,in\,\,S\,\,such\,that\,\,
l(x_{1}) = l(x_{2})\,\,and x_{1} \neq x_{2}.\\
we\,can\,see\,that\,l(000)=l(111)=3\,\,and\,000\neq 111 \\
so\,l\,is\,not\,one-to-one\\
b-\\
l: S \rightarrow \mathbb{Z}^{nonneg}\,\,is\,\,onto\,\Leftrightarrow \,\\
\forall y\,in\,\mathbb{Z}^{nonneg} ,\exists x \in S\,such\,that\, l(x) = y.\\
proof:\\
if\,y=0\,there\,is\,\xi (empty\,string)\,such\,that\,l(\xi )=0\\
if\,y=n> 0\\
let\,s=1111...1\,(n\,times.)\,such\,that\,l(1111...1 )=n \\
so\,l\,is\,onto
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