(i) $f(x)=|x+2|-1$

We know that $|x+2| \geq 0$ for every $x \in \mathbf{R}$.

Therefore, $f(x)=|x+2|-1 \geq-1$ for every $x \in \mathbf{R}$.

The minimum value of $f$ is attained when $|x+2|=0$.

$|x+2|=0$

$\Rightarrow x=-2$

$\therefore$ Minimum value of $f=f(-2)==|-2+2|-1=-1$

Hence, function f does not have a maximum value.

(ii) $g(x)=-|x+1|+3$

We know that $-|x+1| \leq 0$ for every $x \in \mathbf{R}$.

Therefore, $g(x)=-|x+1|+3 \leq 3$ for every $x \in \mathbf{R}$.

The maximum value of $g$ is attained when $|x+1|=0$.

$|x+1|=0$

$\Rightarrow x=-1$

$\therefore$ Maximum value of $a=a(-1)=-|-1+1|+3=3$

(iii) $h(x)=\sin 2 x+5$

We know that $-1 \leq \sin 2 x \leq 1$.

$\Rightarrow-1+5 \leq \sin 2 x+5 \leq 1+5$

$\Rightarrow 4 \leq \sin 2 x+5 \leq 6$

Hence, the maximum and minimum values of h are 6 and 4 respectively.

(iv) $f(x)=|\sin 4 x+3|$

We know that $-1 \leq \sin 4 x \leq 1$.

$\Rightarrow 2 \leq \sin 4 x+3 \leq 4$

$\Rightarrow 2 \leq|\sin 4 x+3| \leq 4$

Hence, the maximum and minimum values of f are 4 and 2 respectively.

(v) $h(x)=x+1, x \in(-1,1)$

Here, if a point $x_{0}$ is closest to $-1$, then we find $\frac{x_{0}}{2}+1>x_{0}+1$ for all $x_{0} \in(-1,1)$.

Also, if $x_{1}$ is closest to 1 , then $x_{1}+1<\frac{x_{1}+1}{2}+1$ for all $x_{1} \in(-1,1)$.

Hence, function h(x) has neither maximum nor minimum value in (−1, 1).