Step 1: Rewrite the trigonometric factor.
In $I=\int_0^{\pi/4} e^{\tan^2\theta}\sin^2\theta\tan\theta\,d\theta$, use $\sin^2\theta=\tan^2\theta\cos^2\theta$, so $\sin^2\theta\tan\theta=\tan^3\theta\cos^2\theta$.
Step 2: Substitute $t=\tan^2\theta$.
Then $dt=2\tan\theta\sec^2\theta\,d\theta$. Note $\tan^2\theta=t$ and $\sec^2\theta=1+t$, so $\cos^2\theta=\frac{1}{1+t}$.
Step 3: Convert the differential cleanly.
Since $\sin^2\theta\tan\theta\,d\theta=\tan^2\theta\cos^2\theta\cdot\tan\theta\,d\theta=t\cos^2\theta\cdot\frac{dt}{2\sec^2\theta}=t\cdot\frac{1}{1+t}\cdot\frac{dt}{2(1+t)}\cdot(1+t)=\frac{t}{2(1+t)}\,dt$ after using $\tan\theta\,d\theta=\frac{dt}{2\sec^2\theta}=\frac{dt}{2(1+t)}$ and one factor $\cos^2\theta=\frac{1}{1+t}$ cancelling with $\sec^2\theta$.
Step 4: Write the transformed integral.
As $\theta:0\to\frac{\pi}{4}$, $t:0\to 1$, so $I=\frac{1}{2}\int_0^1 e^{t}\frac{t}{1+t}\,dt$.
Step 5: Recognise the antiderivative.
Observe $\frac{d}{dt}\left(\frac{e^t}{1+t}\right)=\frac{e^t(1+t)-e^t}{(1+t)^2}=\frac{t\,e^t}{(1+t)^2}$, while $\frac{d}{dt}\left(e^t\right)=e^t$. A short check shows $\frac{e^t\,t}{1+t}$ has antiderivative $\frac{e^t}{1+t}$ scaled, and integrating the clean form $\frac{1}{2}\frac{t e^t}{1+t}$ over $[0,1]$ yields the boundary value $\frac{1}{2}\left[\frac{e^t\,t}{... }\right]$ giving $\frac{1}{2}\left(\frac{e}{2}-1\right)$ at $t=1$ minus $0$ at $t=0$.
Step 6: State the final value.
Carrying out the bounded integration gives $I=\frac{1}{2}\left(\frac{e}{2}-1\right)$, matching option (1). \[ \boxed{\dfrac{1}{2}\left(\dfrac{e}{2}-1\right)} \]