(A) Let $y=\frac{x^2+2 x+4}{x+2}$
$$
\begin{aligned}
& \Rightarrow x^2+(2-y) x+(4-2 y)=0 \\
& \Rightarrow \quad(2-y)^2-4(4-2 y) \geq 0 \\
& \Rightarrow \quad y^2+4 y-12 \geq 0 \\
& \Rightarrow \quad y \leq-6, y \geq 2 \\
&
\end{aligned}
$$
$\therefore$ Minimum value of $y$ is 2 .
(B) Since, $(A+B)(A-B)=(A-B)(A+B)$
$$
\begin{aligned}
& \Rightarrow \quad A^2-A B+B A-B^2=A^2+A B-B A-B^2 \\
& \Rightarrow \quad A B=B A \\
& \text { and } \quad(A B)^t=(-1)^k A B \\
& \Rightarrow \quad B^t A^t=(-1)^k A B \\
& \Rightarrow \quad-B A=(-1)^k A B \quad\left[\because B^t=-B, A^t=A\right] \\
& \Rightarrow \quad B A=(-1)^{k+1} A B \\
& \Rightarrow \quad(-1)^{k+1}=1 \\
&
\end{aligned}
$$

$\therefore k+1$ is even or $k$ is odd.
(C) $1 < 2^{\left(-k+3^{-a}\right)} < 2 \Rightarrow 0 < -k+3^{-a} < 1$
Given, $a=\log _3 \log _3 2 \Rightarrow 3^a=\log _3 2$
$$
\begin{array}{lc}
\Rightarrow & 3^{-a}=\log _2 3 \\
\therefore & k < \log _2 3 < 2 \\
\text { and } & 1+k>\log _2 3>1 \Rightarrow k>0
\end{array}
$$
From Eqs. (ii) and (iii), $0 < k < 2 \Rightarrow k=1$
$[\because k$ is an integer]
$$
\begin{array}{rlrl}
\text { (D) } & \sin \theta & =\cos \phi \\
\Rightarrow & \cos \left(\frac{\pi}{2}-\theta\right) & =\cos \phi \\
\Rightarrow & \frac{\pi}{2}-\theta & =2 n \pi \pm \phi, n \in Z \\
\Rightarrow & & \theta \pm \phi-\frac{\pi}{2} & =-2 n \pi, n \in Z \\
\Rightarrow & & \frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right) & =-2 n, n \in Z
\end{array}
$$