a. Let $a * a = b$. $(a * a) * a = b*a$. Since $(A, *)$ is a semigroup, $*$ is closed and associative. So, $(a *a) * a = a * (a*a) \implies a * b = b * a$, which is possible only if $a = b$. Thus we proved $a * a = a.$
b. Let $(a * b)*a = c \\ \implies (a * b) * a * a = c * a \\ \implies a * b * a = c * a \\ \implies c * a = a.$
Similarly, $a * (a * b * a ) = a * c \\ \implies a * a * (b * a) = a * c \\ \implies a * (b * a) = a * c \\ \implies a * c = a = c * a.$
So, $c = a$.
c. Let $(a * b)*c = d. \\ \implies (a * b) * c * c = d * c \\ \implies a * b * c = d * c \\ \implies d * c = d.$
Similarly, $a * (a * b * c ) = a * d \\ \implies a * a * (b * c) = a * d \\ \implies a * (b * c) = a * d \\ \implies a * d = d.$
Thus $d * c = a * d = d $
Now $c * d *c = c * a * d = c * d \\\implies c = c * a * d = c * d$
and
$d * c * a = a * d * a = d * a \\\implies d * c * a = a = d * a$
So,
$a * c = (d*a)*(c*d) \\= d*(a*c)* d = d.$
Thus, $a*b*c = a * c.$