Answer
\begin{align}
&(a) F(0) = 0 \\
&(b) F'(0) = \frac{1}{5} \\
&(c) F''(0) = -\frac{3}{25}
\end{align}
Work Step by Step
$\text {The given function is}$
\begin{align}
F(x) = \int_{0}^{x}\frac{\cos{t}}{t^{2}+3t+5} \ dt
\end{align}
$ \text {(a)}$
\begin{align}
F(0) = \int_{0}^{0}\frac{\cos{t}}{t^{2}+3t+5} \ dt = 0
\end{align}
$\text {(b)} $
\begin{align}
F'(x) = \frac{\cos{x}}{x^{2}+3x+5}
\end{align}
$\text {Thus,} $
\begin{align}
F'(0) = \frac{\cos{x}}{x^{2}+3x+5} = \frac{1}{5}
\end{align}
$\text {(c)}$
\begin{align}
F''(x) = -\frac{\sin {x} \times(x^{2}+3x+5)+ \cos{x} \times (2x+3)}{(x^{2}+3x+5)^2}
\end{align}
$\text {Thus,} $
\begin{align}
F''(0) = -\frac{\sin {x} \times(x^{2}+3x+5)+ \cos{x} \times (2x+3)}{(x^{2}+3x+5)^2} = -\frac{3}{25}
\end{align}
