Answer
$$\eqalign{
& \left( {\text{a}} \right)0 \cr
& \left( {\text{b}} \right)0 \cr
& \left( {\text{c}} \right)1 \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( x \right) = \int_0^x {\frac{{\sin t}}{{{t^2} + 1}}dt} \cr
& \left( {\text{a}} \right)F\left( 0 \right) = \int_0^0 {\frac{{\sin t}}{{{t^2} + 1}}dt} = 0 \cr
& {\text{Differentiate}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {\int_0^x {\frac{{\sin t}}{{{t^2} + 1}}dt} } \right] = \frac{{\sin x}}{{{x^2} + 1}} \cr
& \left( {\text{b}} \right)F'\left( 0 \right) = \frac{{\sin 0}}{{{0^2} + 1}} = 0 \cr
& F''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{\sin x}}{{{x^2} + 1}}} \right] \cr
& F''\left( x \right) = \frac{{\left( {{x^2} + 1} \right)\left( {\cos x} \right) - \sin x\left( {2x} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \left( {\text{c}} \right)F''\left( 0 \right) = \frac{{\left( {{0^2} + 1} \right)\left( {\cos 0} \right) - \sin 0\left( {2\left( 0 \right)} \right)}}{{{{\left( {{0^2} + 1} \right)}^2}}} = 1 \cr} $$
