(a) f(0)=0 and f(3)=9, and 0<V<9 so 0<c<3.
Let's go for c=1.5 the centre of the interval: f(1.5)=2.25>V when V=2.
Let's now go for c=1.4 in an effort to "pinch" the root: f(1.4)=1.96<V, so 1.4<c<1.5.
We have c correct to 1 dec place. Let's go for the 2nd dec place by putting c=1.45; f(1.45)=2.1025>V and we already know that f(1.4)<V; we now know 1.4<c<1.45, so let's try an intermediate value c=1.42, f(1.42)>V. And now put c=1.41 to pinch the root: f(1.41)<V so we know 1.41<c<1.42. Both values are within [0,3] so we have the upper and lower limits for c. We don't have to consider negative values for c because the interval is positive in scope.
(b) f(x)=sin(x) on [0,π/2]. sin(0)=0 and sin(π/2)=1. V=1/2=0.5, which is halfway in the range so let's put c=π/4. f(π/4)=0.7071>V; try c=π/8, f(π/8)=0.3827<V. Now try c=π/6 which lies between π/4 and π/8, f(π/6)=0.5=V. So we have the exact root c=π/6.