Step 1: Understand the shape.
$f(x)=|x+1|e^{-x^2}$ is always nonnegative, touches $0$ at $x=-1$, and the $e^{-x^2}$ factor pulls everything toward $0$ far from the origin.
Step 2: Locate the global minimum first.
Since $f(x)\ge 0$ and $f(-1)=0$, the global minimum is at $x=-1$, with value $0$. This already kills options 3 and 4, which place a minimum in $(0,1)$.
Step 3: Split off the modulus to hunt maxima.
For $x<-1$, $f(x)=-(x+1)e^{-x^2}$. Differentiate to find turning points in $(-2,-1)$.
Step 4: Solve the critical equation.
Setting the derivative to zero leads to $2x^2+2x-1=0$, so $x=\frac{-1\pm\sqrt 3}{2}$. The root $x=\frac{-1-\sqrt 3}{2}\approx -1.37$ lies in $(-2,-1)$.
Step 5: Confirm it is a maximum.
The derivative switches from positive to negative there, so this critical point is a local maximum, and it sits inside $(-2,-1)$.
Step 6: Reject the rest.
No maximum lies in $(1,2)$, so option 2 fails; the minimum is at $x=-1$, not in $(0,1)$. Hence the true statement is option 1.
\[ \boxed{f \text{ has point of maxima in } (-2,-1)} \]