The basic identity is the first-return decomposition: splitting a return at time $n$ according to the time $k$ of the first return,

$p_{ii}^{(n)}=\sum_{k=1}^{n}f_{ii}^{(k)}\,p_{ii}^{(n-k)}, \qquad p_{ii}^{(0)}=1.$

Periodicity with period $2$ means returns are possible only at even times, so

$p_{ii}^{(n)}=0 \text{ and } f_{ii}^{(n)}=0 \quad\text{for every odd } n .$

(III) is immediate: $7$ is odd, so $f_{ii}^{(7)}=0$ . True.

(I). Take $n=2$ in $(\star)$ :

$p_{ii}^{(2)}=f_{ii}^{(1)}p_{ii}^{(1)}+f_{ii}^{(2)}p_{ii}^{(0)} =f_{ii}^{(1)}p_{ii}^{(1)}+f_{ii}^{(2)} ,$

which rearranges to exactly $f_{ii}^{(2)}=p_{ii}^{(2)}-p_{ii}^{(1)}f_{ii}^{(1)}$ . True (in fact for any chain, periodic or not; here both subtracted terms happen to be $0$ ).

(II). Take $n=4$ in $(\star)$ :

$p_{ii}^{(4)}=f_{ii}^{(1)}p_{ii}^{(3)}+f_{ii}^{(2)}p_{ii}^{(2)} +f_{ii}^{(3)}p_{ii}^{(1)}+f_{ii}^{(4)} .$

Periodicity kills the odd-indexed terms: $p_{ii}^{(1)}=p_{ii}^{(3)}=0$ and $f_{ii}^{(1)}=f_{ii}^{(3)}=0$ . What survives is

$p_{ii}^{(4)}=f_{ii}^{(2)}p_{ii}^{(2)}+f_{ii}^{(4)} \quad\Longrightarrow\quad f_{ii}^{(4)}=p_{ii}^{(4)}-p_{ii}^{(2)}f_{ii}^{(2)} . \quad\checkmark$

All three statements hold.