If $\tan 2\alpha \cdot \tan \alpha = 1$, then what is $\alpha$? Different methods give different answers. 
If $\tan 2\alpha\cdot\tan \alpha = 1$, then what is $\alpha$?

I tried two methods but got two different answers.
Method 1:
$$\begin{align}
\tan 2\alpha\cdot\tan \alpha = 1
&\implies
\frac{2\tan \alpha}{1 - \tan ^2 \alpha}\;\tan \alpha = 1 \tag{1a}\\[6pt]
&\implies 2\tan ^2\alpha = 1 - \tan ^2\alpha \tag{1b}\\[6pt]
&\implies \tan ^2 \alpha = \frac{1}{3} \tag{1c}\\[6pt]
&\implies \tan \alpha = \pm\frac{1}{\sqrt 3} \tag{1d}\\[6pt]
&\implies \tan \alpha = \tan\left(\pm\frac{\pi}{6}\right) \tag{1e}\\[6pt]
&\implies \alpha = n\pi \pm \frac{\pi}{6} \;\text{where}\; n \in \mathbb{Z} \tag{1f}
\end{align}$$
Method 2:
$$\begin{align}
\tan 2\alpha \cdot \tan \alpha = 1
&\implies \tan 2\alpha = \frac{1}{\tan \alpha} \tag{2a}\\[6pt]
&\implies \tan 2\alpha = \cot \alpha \tag{2b}\\[6pt]
&\implies \tan 2\alpha = \tan\left(\frac{\pi}{2} - \alpha\right) \tag{2c}\\[6pt]
&\implies 2\alpha = n\pi + \frac{\pi}{2} - \alpha \text{?} \tag{2d}\\[6pt]
&\implies \alpha = \frac{1}{3}\left(n\pi + \frac{\pi}{2}\right)\;\text{where}\; n \in \mathbb{Z} \tag{2e}
\end{align}$$
Which one is correct? Is there any mistake in the above solutions?

Your first method is correct.
In the second method, notice that the equation
$$\tan 2\alpha = \frac{1}{\tan\alpha}$$
is not valid when $\alpha = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$ or when $\alpha = \frac{\pi}{4} + \frac{n\pi}{2}, n \in \mathbb{Z}$ or when $\alpha = n\pi, n \in \mathbb{Z}$.  Therefore, in your solution 
$$\alpha = \frac{\pi}{6} + \frac{n\pi}{3}, n \in \mathbb{Z}$$
$n \neq 3k + 1$, $k \in \mathbb{Z}$ since that would imply
$$\alpha = \frac{\pi}{6} + k\pi + \frac{\pi}{3} = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
which is not valid.  
If we replace $n$ by $3k$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
If we replace $n$ by $3k - 1$, we obtain
$$\alpha = \frac{\pi}{6} + k\pi - \frac{\pi}{3} = -\frac{\pi}{6} + k\pi, k \in \mathbb{Z}$$
which agrees with your first solution.
