Answer
a) $\color{blue}{F'_Y(y) =\dfrac{e^{-y}}{(1+e^{-y})^2} \gt 0,\ -\infty\lt y\lt\infty},$ so $F_Y(y)$ is increasing on $-\infty\lt y\lt\infty$.
b) $\color{blue}{\lim\limits_{y\ \to\ -\infty}\ F_Y(y)= \dfrac{1}{\infty} = 0}$.
c) $\color{blue}{\lim\limits_{y\ \to\ \infty}\ F_Y(y)= \dfrac{1}{1+0} = 1}$.
d) $\color{blue}{f_Y(y) = \dfrac{e^{-y}}{(1+e^{-y})^2},\ -\infty\lt y\lt \infty}.$
Work Step by Step
a) $\underline{F_Y(y)\ \text{is increasing}}$
Observe that $\dfrac{dF_Y(y)}{dy} = \dfrac{d}{dy}\left(\dfrac{1}{1+e^{-y}}\right) $ $= \dfrac{d}{dy}(1+e^{-y})^{-1} = -(1+e^{-y})^{-2}(-e^{-y})$ $= \dfrac{e^{-y}}{(1+e^{-y})^2} \gt 0,\ y\in \mathbb{R},$ where we have used the Chain Rule to differentiate.
Thus, $\color{blue}{F'_Y(y) =\dfrac{e^{-y}}{(1+e^{-y})^2} \gt 0,\ -\infty\lt y\lt\infty}$, so that $F_Y(y)$ is increasing for $-\infty\lt y\lt \infty$, i.e., since the frist derivative (with respect to $y$) of $F_Y(y)$ is positive on the entire real line, then $F_Y(y)$ must be increasing on the entire real line.
b) $\underline{\lim\limits_{y\ \to\ -\infty}\ F_Y(y) = 0}$
$\begin{align*}
\lim\limits_{y\ \to\ -\infty}\ F_Y(y) &= \lim\limits_{y\ \to\ -\infty}\ \frac{1}{1+e^{-y}} \\
&= \frac{1}{1+e^{-(-\infty)}} \\
&= \frac{1}{1+e^\infty} \\
&= \frac{1}{1+\infty} \\
&= \frac{1}{\infty} \\
\color{blue}{\lim\limits_{y\ \to\ -\infty}\ F_Y(y)} &\color{blue}{=0}.
\end{align*}$
c) $\underline{\lim\limits_{y\ \to\ \infty}\ F_Y(y) = 1}$
$\begin{align*}
\lim\limits_{y\ \to\ \infty}\ F_Y(y) &= \lim\limits_{y\ \to\ \infty}\ \frac{1}{1+e^{-y}} \\
&= \frac{1}{1+e^{-\infty}} \\
&= \frac{1}{1+e^\infty} \\
&= \frac{1}{1+0} \\
&= \frac{1}{1} \\
\color{blue}{\lim\limits_{y\ \to\ \infty}\ F_Y(y)} &\color{blue}{=1}.
\end{align*}$
d) $\underline{\text{Associated pdf}\ f_Y(y)}$
Since the associated pdf $f_Y(y) = \dfrac{dF_Y(y)}{dy} = F_Y'(y)$, then from part a) above, $\color{blue}{f_Y(y) = \dfrac{e^{-y}}{(1+e^{-y})^2},\ -\infty\lt y\lt \infty}.$
