= Solution
Suppose $A=\operatorname{ran}f$ for a <partial computable function> $f$. Since $A\ne\varnothing$, fix $a_0\in A$. Decode each input $w$ as a pair $((w)_0,(w)_1)=(u,t)$. Using the <truncated computation function>, define
$$
h(w)=
\begin{cases}
f(u),&\text{if the computation of $f(u)$ halts within $t$ steps},\\
a_0,&\text{otherwise}.
\end{cases}
$$
This is a <total computable function>, and every output lies in $A$. Conversely, if $a=f(u)$, choosing $t$ at least the halting time gives an input $w$ with $h(w)=a$. Thus $\operatorname{ran}h=A$, proving (iii)$\Rightarrow$(iv).
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