One year ago, a man was 8 times as old as his son.

Solution:

(i)
Let the present age of the son be x years.

∴ Present age of the man = x2 years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x2 − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$\therefore x^{2}-1=8(x-1)$

$\Rightarrow x^{2}-1=8 x-8$

$\Rightarrow x^{2}-8 x+7=0$

$\Rightarrow x^{2}-7 x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow(x-1)(x-7)=0$

$\Rightarrow x-1=0$ or $x-7=0$

$\Rightarrow x=1$ or $x=7$

∴ x = 7                (Man's age cannot be 1 year) 

Present age of the son = 7 years

Present age of the man = 72 years = 49 years

(ii)
Let the age of man be m and the age of son be s

It is given that man is $3 \frac{1}{2}$ times as old as his son.

$\Rightarrow m=3 \frac{1}{2} s$

$\Rightarrow m=\frac{7}{2} s \quad \ldots$ (i)

Also given that 

$m^{2}+s^{2}=1325 \quad \ldots \ldots$ (ii)

Put value of (i) in (ii), we get

$\left(\frac{7}{2} s\right)^{2}+s^{2}=1325$

$\frac{\Rightarrow 49 s^{2}+4 s^{2}}{4}=1325$

$53 s^{2}=5300$

$\Rightarrow s^{2}=100$

$\Rightarrow s=\pm 10$

Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have

$m=\frac{7}{2} s$

$\Rightarrow m=\frac{7}{2} \times 10$

$\Rightarrow m=35$

So, age of man = 35 years
Age of son = 10 years