1. First simplify the complex number:

$\frac{1+i}{1-i}$

Multiply numerator and denominator by the conjugate of the denominator, $1+i$:

$\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}=\frac{(1+i)^2}{1-i^2}$

Now,

$(1+i)^2 = 1+2i+i^2 = 2i$

and

$1-i^2 = 1-(-1)=2$

So,

$\frac{1+i}{1-i}=\frac{2i}{2}=i$

Hence the given condition becomes

$i^{m/2}=1 \quad \text{and} \quad i^{n/3}=1$

2. Use the fact that powers of $i$ repeat with period $4$:

$i^k=1 \iff k \text{ is a multiple of } 4$

So for the first equation,

$i^{m/2}=1 \implies \frac{m}{2}=4r \quad (r\in \mathbb{N}_0)$

Thus,

$m=8r$

The least natural value of $m$ is

$m=8$

3. For the second equation,

$i^{n/3}=1 \implies \frac{n}{3}=4s \quad (s\in \mathbb{N}_0)$

Thus,

$n=12s$

The least natural value of $n$ is

$n=12$

4. Now compute the greatest common divisor:

$\gcd(8,12)=4$

Therefore, the required answer is

$\boxed{4}$