The height of a cone increases by k%,

Let h be the height, y be the surface area, V be the volume, l be the slant height and r be the radius of the cone.

Let $\Delta h$ be the change in the height, $\Delta r$ be the change in the radius of base and $\Delta l$ be the change in the slant height. Semi - vertical angle ramaining the same.

$\therefore \frac{\Delta h}{h}=\frac{\Delta r}{r}=\frac{\Delta l}{l}$

Also, $\frac{\Delta h}{h} \times 100=k$

Then, $\frac{\Delta h}{h} \times 100=\frac{\Delta r}{r} \times 100=\frac{\Delta l}{l} \times 100=k$        ......(1)

(i) Total surface area of the cone, $T=\pi r l+\pi r^{2}$

Differentiating both sides w.r.t. $r$, we get

$\frac{d T}{d r}=\pi l+\pi r \frac{d l}{d r}+2 \pi r$

$\Rightarrow \frac{d T}{d r}=\pi l+\pi r \frac{l}{r}+2 \pi r$       $\left[\right.$ From $\left.(1), \frac{d l}{d r}=\frac{\Delta l}{\Delta r}=\frac{l}{r}\right]$

$\Rightarrow \frac{d T}{d r}=\pi l+\pi l+2 \pi r$

$\Rightarrow \frac{d T}{d r}=2 \pi(l+r)$

$\therefore \Delta T=\frac{d T}{d r} \Delta r=2 \pi(l+r) \times \frac{k r}{100}=\frac{2 k r \pi(l+r)}{100}$

$\therefore \frac{\Delta T}{T} \times 100=\frac{\left(\frac{2 k r \pi(l+r)}{100}\right)}{2 \pi r(l+r)} \times 100=2 k \%$

Hence, the percentage increase in total surface area of cone is $2 k$.

(ii) Volume of cone, $V=\frac{1}{3} \pi r^{2} h$

Differentiating both sides w.r.t. $h$, we get

$\frac{d V}{d h}=\frac{1}{3} \pi r^{2}+\frac{1}{3} \pi h 2 r \frac{d r}{d h}$

$\Rightarrow \frac{d V}{d h}=\frac{1}{3} \pi r^{2}+\frac{1}{3} \pi h 2 r \frac{r}{h}$      $\left[\operatorname{From}(1), \frac{d r}{d h}=\frac{\Delta r}{\Delta h}=\frac{r}{h}\right]$

$\Rightarrow \frac{d V}{d h}=\frac{1}{3} \pi r^{2}+\frac{2}{3} \pi r^{2}$

$\Rightarrow \frac{d V}{d h}=\pi r^{2}$

Hence, the percentage increase in the volume of the cone is $3 k$.