Function -> HeELP

1) Let for each x in [0,1],
F(x) = \int_0^x f(t)\ \mathrm dt
and further define a function
g(x) = e^{-x}F(x)\quad x\in[0,1]
Since f(x) is continuous in [0,1], so F(x) is differentiable and hence g(x) is differentiable in (0,1). At the same time g(0) = g(1) =0. Hence according to Rolle's theorem there exist at least one c in (0,1) so that g'(c) =0.
But
g'(x) = e^-x F'(x) - e^-x F(x) = e^-x (F'(x) - F(x))
Since e^-x ≠0, therefore there is atleast one c in (0,1) such that
F'(c) = F(c)
i.e.
f(c) = \int_0^c f(t)\mathrm dt

For 2). The given equation is equivalent to
\int_0^\alpha (\cos (x+\alpha^2) - \cos x)\mathrm dx =0
i.e.
-2\sin(\alpha^2/2)\int_0^\alpha \sin\left(x+\frac{\alpha^2}{2}\right)\mathrm dx = 0
This gives either
\sin \left(\frac{\alpha^2}{2}\right) = 0 ---- \quad (1)
or
\int_0^\alpha \sin\left(x+\frac{\alpha^2}{2}\right)\mathrm dx = 0 ---\quad (2)
From (1), we get
\alpha^2 = 2n \pi for some integer n.
Since \alpha lies between 2 and 3, its square will lie between 4 and 9 and so the only possibility is that \alpha^2 = 2 \pi giving
\boxed{\alpha = \sqrt{2\pi}}
From (2) we get
\cos\left(\alpha + \frac{\alpha^2}{2}\right) = \cos \left(\frac{\alpha^2}{2}\right)
So that
\alpha + \frac{\alpha^2}{2} = 2k\pi \pm \frac{\alpha^2}{2}
for integer k.
Taking the plus sign, we get \alpha = 2k\pi which cannot be true for any integral k (since alpha lies between 2 and 3).
Taking the minus sign, we get
\alpha + \alpha^2 = 2k\pi
Since 6<\alpha + \alpha^2 < 12, the only possibility is that
\alpha + \alpha^2 = 2\pi i.e. \alpha^2 + \alpha -2\pi = 0 which has got one solution in (2,3) which is
\alpha = \frac{1}{2}\left(\sqrt{1+8\pi}-1\right)
And hence the original equation has exactly two roots in (2,3).

3) For the given interval x-[x] = x and hence the given integral is simply
\int_0^{\pi/4} (\tan^nx + \tan^{n-2}x)\mathrm dx
which is quite trivially equal to
\frac{1}{n-1}
for n≠1.