I'll try from what I understand.
(a) f(x)=x^2 on [0,3]; f(0)=0, f(3)=9; f(c)=V=2. So since 2 is within the limits, there must be a value c. On the given interval c would be the irrational number √2. If we put c=1.41, f(c)=1.9881 and if we put c=1.42, f(c)=2.0164. So we know by the IVT that c is between these two values: c=1.41, c=1.42.
(b) f(0)=0 and f(π/2)=1. V=1/2 lies between 0 and 1. We want f(c)=1/2 so let's see what happens if we take c halfway between 0 and π/2, so c=π/4 and f(π/4) can be found by looking at a right-angled isosceles triangle of unit side length. Then sin c = 1/√2 or √2/2=0.71 approximately. 0.71 > 0.5 (=V), so we now know that c < π/4.
So c is between 0 and π/4 according to IVT. Try c=π/8 then f(π/8)=0.38 which is less than 0.5, so c must be between π/8 and π/4. Since π is irrational but equivalent to 180 degrees, we could use degrees as the unit instead of radians. Then π/8=180/8=22.5 degrees. So c is between 22.5 and 45 degrees. The next angle we can try is 30 degrees because an equilateral triangle can be bisected into two right-angled triangles. The bisected vertex angle of a unit equilateral triangle is 30° and the sine must be 1/2, which is V, because the side opposite the angle is half the length of the side. So c=30°, which converts to 30π/180=π/6. So c=π/6.
f(x)=x^2-2 on [0,3]; f(0)=-2 and f(3)=7. There is a change of sign where x=c, the root of the expression. The midpoint of the interval is 1.5 when f(1.5)=0.25. This is positive so c<1.5. Next try the midpoint of [0,1.5]=0.75. f(0.75)<0. Therefore c is in the interval [0.75,1.5]=1.125; f(1.125)<0 so c is in [1.125,1.5]. If we reduce this to one dec place we have [1.1,1.5], the midpoint of which is 1.3; f(1.3)<0, so we try [1.3,1.5]=1.4 as midpoint. f(1.3)=-0.31 and f(1.5)=0.25. And f(1.4)=-0.04 which rounds up to 0. So the root is 1.4.