$I_0$

$S' \rightarrow .S , \$$

$S \rightarrow . A A, \$$

$A \rightarrow .aA, a/b$

$\ \ \ \ \ \ \ \ \ \ .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? $\epsilon a/ b$