See the definitions from Q7. ,Q11. and Q13.
$(i)$ $5 \ast 7 = L.C.M (5,7) = 35$ and $20 \ast 16 = L.C.M(20,16) = 80$.
$(ii)$ Yes, $\ast$ is commutative. As we know that $L.C.M(a,b) = L.C.M(b,a)$ for any two natural number in $N$.
$\implies \, a \ast b = b \ast a \, \forall a,b \in N$.
$(iii)$ Yes, $\ast$ is associative. As we know that $\,\,\,\,L.C.M(a, L.C.M(b, c)) = L.C.M(L.C.M(a,b), c)$
$\implies a \ast (b \ast c) = (a \ast b) \ast c \, \forall a,b,c \in N$.
$(iv)$ The element $e = 1$ in $N$ is the identity element, because $a \ast 1 = 1 \ast a = a, \, \forall a \in N$.(as we know that $L.C.M(a,1) = a$).
$(v)$ No element other then $identity \, element \, 1$ is invertible.
Proof: If possible let $a' \in N$ be the inverse of $a \in N$. Then $a \ast a' = a' \ast a = 1\, \implies \, L.C.M(a,a') = 1$ which is not possible for any element other then $a = a' = 1$. So, $a \neq 1 $ in $N$ is not invertible.