Let f: ℝ → ℝ be a twice differentiable function such that f''(x) >...

Given $f''(x) > 0$ for all $x \in \mathbb{R}$, which means $f'(x)$ is a strictly increasing function.
We are given $f'(a-1) = 0$. Since $f'(x)$ is strictly increasing, $f'(x) < 0$ for $x < a-1$ and $f'(x) > 0$ for $x > a-1$.
The function $g(x)$ is defined as $g(x) = f(\tan^2 x - 2\tan x + a)$ for $x \in (0, \pi/2)$.
Let $u(x) = \tan^2 x - 2\tan x + a = (\tan x - 1)^2 + a - 1$.
Differentiating $g(x)$ with respect to $x$:
$g'(x) = f'(u(x)) \cdot u'(x) = f'((\tan x - 1)^2 + a - 1) \cdot (2\tan x \sec^2 x - 2\sec^2 x)$
$g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2\sec^2 x (\tan x - 1)$.

Case 1: $x \in (0, \pi/4)$.
In this interval, $0 < \tan x < 1$, so $(\tan x - 1) < 0$.
Also, $(\tan x - 1)^2 > 0$, so $u(x) = (\tan x - 1)^2 + a - 1 > a - 1$.
Since $f'(x)$ is increasing and $f'(a-1) = 0$, for $u(x) > a - 1$, we have $f'(u(x)) > 0$.
Thus, $g'(x) = (+)(+)(-) < 0$. So $g(x)$ is decreasing in $(0, \pi/4)$.
Statement (I) is False.

Case 2: $x \in (\pi/4, \pi/2)$.
In this interval, $\tan x > 1$, so $(\tan x - 1) > 0$.
Also, $(\tan x - 1)^2 > 0$, so $u(x) = (\tan x - 1)^2 + a - 1 > a - 1$.
Thus, $f'(u(x)) > 0$.
Then $g'(x) = (+)(+)(+) > 0$. So $g(x)$ is increasing in $(\pi/4, \pi/2)$.
Statement (II) is False.

Therefore, neither (I) nor (II) is true.