$I_0$
$S' \rightarrow .S , \$$
$S \rightarrow . A A, \$$
$A \rightarrow .aA, a/b$
$\ \ \ \ \ \ \ \ \ \ .b,a / b $
We begin calculating closure of $S' \rightarrow .S$
How $\$$ is lookahead of $S \rightarrow . A A$?
$\epsilon \ \$$ is after S in prodcution $S' \rightarrow .S$ so what is first($)?
It is $\$$, which is lookahead of $S \rightarrow . A A, \$$
How are $a/b$ lookaheads of $A \rightarrow .aA, a/b$ and $A \rightarrow .b,a / b ?$
Productions of A came because of $S \rightarrow . A A, \$$ and what is after A?
A$, So first of A = {a,b} Hence, we got lookaheads a,b
You have confusion in $I_2$ and $I_3$
$\text{Goto}(I_0,A) \rightarrow I_2$
$I_2$
$S \rightarrow A.A, \$$
$A \rightarrow .aA, \$$
$A \rightarrow .b, \$$
How $\$$ is looahead for $A \rightarrow .aA, \$$ and $A \rightarrow .b, \$$?
A's production came because of $S \rightarrow A.A, \$$ and what is after A?
$\$$ so First$(\$)$ = ${\$}$
Now coming to $I_3$
$\text{GOTO}(I_0,a) \rightarrow I_3$
$I_3$
$A \rightarrow a.A , a/b$
$\color{red}{A \rightarrow .aA , a/b}$
$\color{red}{A \rightarrow .b , a/b}$
Why $a/b$ are lookaheads of $\color{red}{A \rightarrow .aA , a/b}$ and $\color{red}{A \rightarrow .b , a/b}$?
Because these productions came from closure of $A \rightarrow a.A , a/b$ and what is right of A? $\epsilon a/ b$