Given, $f(x)=x^2+7$ and $g(x)=3 x+5$.
(i) $f(3)=(3)^2+7=16$ and $g(-5)=3(-5)+5=-10$
$\therefore f(3)+g(-5)=16-10=6$
(ii) $f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+7=\frac{29}{4}$
and $g(14)=3(14)+5=47$


$$
\therefore f\left(\frac{1}{2}\right) \times g(14)=\frac{29}{4} \times 47=\frac{1363}{4}
$$
(iii) $f(-2)=(-2)^2+7=11$ and $g(-1)=3(-1)+5=2$
$\therefore f(-2)+g(-1)=11+2=13$
(iv) $f(t)=t^2+7$ and $f(-2)=(-2)^2+7=11$
$\therefore f(t)-f(-2)=t^2+7-11=t^2-4$
(v) $f(t)=t^2+7$ and $f(5)=5^2+7=32$
$$
\begin{aligned}
&\therefore \quad \frac{f(t)-f(5)}{t-5}=\frac{t^2+7-32}{t-5} \\
&=\frac{t^2-25}{t-5}=\frac{(t-5)(t+5)}{(t-5)}=t+5 \quad[\therefore t \neq 5]
\end{aligned}
$$