The values of $a$ for which the function $f(x)=\sin x-a x+b$ increases on $R$ are $(-\infty,-1)$.

$$\begin{aligned} \because\quad f^{\prime}(x) & =\cos x-a \\ \text{and}\quad f^{\prime}(x) & >0 \Rightarrow \cos x>a \end{aligned}$$

$$\begin{aligned} &\text { Since, } \quad\cos x \in[-1,1]\\ &\Rightarrow \quad a<-1 \Rightarrow a \in(-\infty,-1) \end{aligned}$$

The function $f(x)=\frac{2 x^2-1}{x^4}$, where $x>0$, decreases in the interval $(1, \infty)$.

$\because \quad f^{\prime}(x)=\frac{x^4 \cdot 4 x-\left(2 x^2-1\right) \cdot 4 x^3}{x^8}=\frac{4 x^5-8 x^5+4 x^3}{x^8}$

$=\frac{-4 x^5+4 x^3}{x^8}=\frac{4 x^3\left(-x^2+1\right)}{x^8}$

Also, $f^{\prime}(x)<0$

$$\begin{array}{lc} \Rightarrow & \frac{4 x^3\left(1-x^2\right)}{x^8}<0 \Rightarrow x^2>1 \\ \Rightarrow & x> \pm 1 \\ \therefore & x \in(1, \infty) \end{array}$$

The least value of function $f(x)=a x+\frac{b}{x}$ (where, $a>0, b>0, x>0)$ is $2 \sqrt{a b}$.

$\because\quad f^{\prime}(x)=a-\frac{b}{x^2}$ and $f^{\prime}(x)=0$

$$\begin{array}{ll} \Rightarrow & a=\frac{b}{x^2} \\ \Rightarrow & x^2=\frac{b}{a} \Rightarrow x= \pm \sqrt{\frac{b}{a}} \end{array}$$

$$\begin{aligned} &\begin{aligned} \text { Now, }\quad & f^{\prime \prime}(x)=-b \cdot \frac{(-2)}{x^3}=+\frac{2 b}{x^3} \\ \text { At } x=\sqrt{\frac{b}{a}},\quad & f^{\prime \prime}(x)=+\frac{2 b}{\left(\frac{b}{a}\right)^{3 / 2}}=\frac{+2 b \cdot a^{3 / 2}}{b^{3 / 2}} \end{aligned} \end{aligned}$$

$=+2 b^{-1 / 2} \cdot a^{3 / 2}=+2 \sqrt{\frac{a^3}{b}}>0 \quad[\because a, b>0]$

$\therefore$ Least value of $f(x), \quad f\left(\sqrt{\frac{b}{a}}\right)=a \cdot \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$

$$\begin{aligned} & =a \cdot a^{-1 / 2} \cdot b^{1 / 2}+b \cdot b^{-1 / 2} \cdot a^{1 / 2} \\ & =\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b} \end{aligned}$$