If we try to solve the first equation

$1 / AC = 0.\overline{0BC}~...~(1)$
$1000 / AC = BC.\overline{0BC}~...~(2)$

$(2) - (1) = 999 / AC = BC$
$999 = AC \times BC$

For the second equation

It's the same as the previous one.

So

We need to solve $999 = AC \times BC$.
As $999 = 3^3 \times 37$ and we need both factors ($AC$ and $BC$) to have exactly $2$ digits, hence there is only one possible answer which is $27 \times 37$.

Thus

$A = 2, B = 3, C = 7$

If we try to solve the first equation

$1 / AC = 0.\overline{0BC}~...~(1)$
$1000 / AC = BC.\overline{0BC}~...~(2)$

$(2) - (1) = 999 / AC = BC$
$999 = AC \times BC$

For the second equation

It's the same as the previous one.

So

We need to solve $999 = AC \times BC$.
As the prime factorization of $999$ is $3^3 \times 37$ and we need both factors ($AC$ and $BC$) to have exactly $2$ digits, hence there is only one possible pair/answer which is $27 \times 37$.

Thus

$A = 2, B = 3, C = 7$ or $A = 3, B = 2, C = 7$