By remainder theorem,

Solution:

(i) Given, $p(x)=x^{3}-2 x^{2}-4 x-1$ and $g(x)=x+1$

Here, zero of $g(x)$ is $-1$.

When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(-1)$

$\therefore \quad p(-1)=(-1)^{3}-2(-1)^{2}-4(-1)-1$

$=-1-2+4-1$

$=4-4=0$

Hence, remainder is $0 .$

(ii) Given, $p(x)=x^{3}-3 x^{2}+4 x+50$ and $g(x)=x-3$

Here, zero of $g(x)$ is 3 .

When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(3)$.

$\therefore \quad p(3)=(3)^{3}-3(3)^{2}+4(3)+50$

$=27-27+12+50=62$

Hence, remainder is 62 .

(iii) Given, $p(x)=4 x^{3}-12 x^{2}+14 x-3$ and $g(x)=2 x-1$

Here, zero of $g(x)$ is $\frac{1}{2}$.

When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p\left(\frac{1}{2}\right)$.

$\therefore \quad p\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{3}-12\left(\frac{1}{2}\right)^{2}+14\left(\frac{1}{2}\right)-3=4 \times \frac{1}{8}-12 \times \frac{1}{4}+14 \times \frac{1}{2}-3$

$=\frac{1}{2}-3+7-3=\frac{1}{2}+1=\frac{1+2}{2}=\frac{3}{2}$

Hence, remainder is $\frac{3}{2}$.

(iv) Given, $p(x)=x^{3}-6 x^{2}+2 x-4$ and $g(x)=1-\frac{3}{2} x$.

Here, zero of $g(x)$ is $\frac{2}{3}$.

When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p\left(\frac{2}{3}\right)$

$\because$ $=\frac{8}{27}-6 \times \frac{4}{9}+2 \times \frac{2}{3}-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4$

$=\frac{8-72+36-108}{27}=\frac{-136}{27}$

Hence, remainder is $\frac{-136}{27}$.