A is correct.
Proving $n \log(a) = \Omega (\log ^2 n)$ is equivalent to proving $\log ^2 n = \text{O}(n) \text{ [a is constant>1]}$
Now, $\log ^2 n = \text{O} (n)$
$\Leftrightarrow 2 \log \log n = \text{O} (\log n) \text{ [taking log on both sides]}$
$\Leftrightarrow \log \log n = \text{O} (\log n)$
which is correct.
B is incorrect.
$\Leftrightarrow (\frac{n}{2}) (\frac{n}{2}+1) (\frac{n}{2}+1) \dots \dots n < n!$
$\Leftrightarrow (\frac{n}{2})^{\frac{n}{2}} < n!$
$\Leftrightarrow n!=(\frac{n}{2})^{\frac{n}{2}}$
C is correct.
Since $U$ and $m$ are constants greatest order term in $= (n+u)^m$ is $n^m$. So $(n+u)^m = n^m$.
D is correct.
$n + \log n < 2n$
$\Leftrightarrow n+ \log (n) = \text{O}(n) \dots (1)$
$n + \log n > W$
$\Leftrightarrow n+ \log (n) = \Omega (n) \dots (2)$
From (1) and (2) $n + \log (n) = \Theta (n)$