Let sum borrowed at 12% p.a. be Rs. x
and that borrowed at 10% p.a. be Rs. (30000 - x)
S.I. at end of 2 years
= Rs. (36480 - 30000)
= Rs. 6480
$$\refore \left( {\frac{{x \times 12 \times 2}}{{100}}} \right) + $$    $$\left[ {\frac{{\left( {30000 - x} \right) \times 10 \times 2}}{{100}}} \right]$$     $$ = 6480$$
$$\eqalign{ & \Leftrightarrow 24x + 600000 - 20x = 648000 \cr & \Leftrightarrow 4x = 48000 \cr & \Leftrightarrow x = 12000 \cr} $$

$$\eqalign{ & {\text{Rate of interest}} = {\text{5}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{sum}} = {\text{Rs}}{\text{. 3200}} \cr & {\text{Time}} = {\text{2 years 6 months}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\frac{1}{2}{\text{years}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{2}{\text{years}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{{\text{Prinicipal}} \times {\text{Time}} \times {\text{Rate}}}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{3200 \times 5 \times 5}}{{2 \times 100}} \cr & \,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}400 \cr & \refore {\text{Amount}} = {\text{Rs}}{\text{. sum}} + {\text{S}}{\text{.I}}. \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {3200 + 400} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. 3600}}{\text{}} \cr} $$

According to question,

Let two sums be Rs. x and Rs. (18750 - x).
Then,
  $$ = x + \frac{{x \times 5 \times 6}}{{100}} = \left( {{\text{18750}} - x} \right) + $$       $$\frac{{\left( {{\text{18750}} - x} \right) \times 5 \times 4}}{{100}}$$
$$\eqalign{ & \Leftrightarrow x + \frac{{30x}}{{100}} = \left( {{\text{18750}} - x} \right) + 3750 - \frac{{20x}}{{100}} \cr & \Leftrightarrow 2x + \frac{x}{2} = 22500 \cr & \Leftrightarrow \frac{{5x}}{2} = 22500 \cr & \Leftrightarrow x = \left( {\frac{{22500 \times 2}}{5}} \right) \cr & \Leftrightarrow x = 9000 \cr} $$
So or sum will be
= ( 18750 - 9000)
= 9750
Hence,
The two sums are Rs. 9000, Rs. 9750

Let time is 't' years and rate of interest is R%
According to question,
$$\eqalign{ & {\bf{Case(I):}} \cr & y = \frac{{x \times {\text{R}} \times {\text{t}}}}{{100}}.....(i) \cr & {\bf{Case(II):}} \cr & z = \frac{{y \times {\text{R}} \times {\text{t}}}}{{100}}.....(ii) \cr} $$
By dividing equation (i) by equation (ii)
$$\eqalign{ & \frac{y}{z} = \frac{{x \times {\text{R}} \times {\text{t}}}}{{y \times {\text{R}} \times {\text{t}}}} \cr & \Rightarrow {y^2} = zx \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{Amount = Rs}}{\text{. 3144}} \cr & {\text{Rate = 8}}\% \cr & {\text{Let, Principal = Rs}}{\text{. }}x \cr & \refore {\text{Time = }} \cr & \frac{{30 + 29 + 31 + 30 + 31 + 30 + 31 + 7}}{{366}} \cr & = \frac{{219}}{{366}} \cr & \refore {\text{SI = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} \cr & \Rightarrow 3144 - x = \frac{{x \times 8 \times 219}}{{100 \times 366}} \cr & = {\text{Rs}}{\text{. 3000}} \cr & \cr & {\bf{Alternate}} \cr & \Rightarrow {\text{P}} + \frac{{{\text{P}} \times 8 \times \frac{{219}}{{365}}}}{{100}} = 3144 \cr & \Rightarrow {\text{P}} + \frac{{{\text{P}} \times 8 \times \frac{3}{5}}}{{100}} = 3144 \cr & \Rightarrow 100{\text{P}} + \frac{{24{\text{P}}}}{5} = 314400 \cr & \Rightarrow \frac{{524{\text{P}}}}{5} = 314400 \cr & \Rightarrow {\text{P = }}\frac{{314400 \times 5}}{{524}} \cr & \Rightarrow {\text{P}} = {\text{600}} \times {\text{5}} \cr & \Rightarrow {\text{P = Rs}}{\text{.}} \, 3000 \cr} $$

$$\eqalign{ & {\text{Let sum be Rs}}{\text{. }}x \cr & {\text{Rate be R}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Time be T years}}{\text{.}} \cr & {\text{Then,}} \cr & \left[ {\frac{{x \times \left( {{\text{R}} \times 2} \right) \times {\text{T}}}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{T}} = 10800\,........(i) \cr & And, \cr & \left[ {\frac{{x \times {\text{R}} \times \left( {{\text{T}} + 2} \right)}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{R}} = 18000\,.......(ii) \cr} $$
Clearly, from (i) and (ii), we cannot find value of x.
So, data is inadequate.

$$\eqalign{ & {\text{P}} = {\text{Rs}}{\text{. 6200}} \cr & {\text{S}}{\text{.I}}{\text{.}} \cr & = {\text{Rs}}{\text{.}}\left( {{\text{9176}} - {\text{6200}}} \right) \cr & = {\text{Rs}}{\text{. }}2976 \cr & {\text{T}} = 4\,{\text{years}} \cr & \refore {\text{Rate}} \cr & = \left( {\frac{{100 \times 2976}}{{6200 \times 4}}} \right)\% \cr & = 12\% \cr & {\text{New }}{\text{rate}} \cr & = \left( {12 + 3} \right)\% \cr & = 15\% \cr & {\text{New}}\,{\text{S}}{\text{.I}}{\text{.}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{{\text{6200}} \times 15 \times 4}}{{100}}} \right) \cr & = \text{Rs.} \,3720 \cr & {\text{New}}\,{\text{amount}} \cr & = {\text{Rs}}{\text{.}}\left( {{\text{6200}} + 3720} \right) \cr & = {\text{Rs}}{\text{.}}\,9920 \cr} $$

Difference between Compound interest and Simple interest for fourth year is Rs. 7280
$$\eqalign{ & P{\left( {\frac{6}{5}} \right)^3}\left( {\frac{1}{5}} \right) - \frac{P}{5} = 7280 \cr & \frac{P}{5}\left[ {{{\left( {\frac{6}{5}} \right)}^3} - 1} \right] = 7280 \cr & \frac{P}{5}\left[ {\frac{{216}}{{125}} - 1} \right] = 7280 \cr & \frac{P}{5} \times \frac{{91}}{{125}} = 7280 \cr & \refore P = 50000 \cr} $$

$$\eqalign{ & {\text{S}}{\text{.I}}{\text{. }}{\text{for 3 years}} \cr & = {\text{Rs}}{\text{.}}\left( {24412.50 - 20925} \right) \cr & = {\text{Rs}}{\text{. }}3487.50 \cr & {\text{S}}{\text{.I}}{\text{. }}{\text{for 2 years}} \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{3487.50}}{3} \times 2} \right) \cr & = {\text{Rs}}{\text{. }}2325 \cr & \refore \text{Principal} \cr & = {\text{Rs}}{\text{.}}\left( {20925 - 2325} \right) \cr & = {\text{Rs}}{\text{. }}18600 \cr & {\text{Hence,}} \cr & {\text{rate}} = \left( {\frac{{100 - 2325}}{{18600 \times 2}}} \right)\% \cr & = 6.25\% \cr} $$