Given:
– We need to find the value of \(\sin^2 10 + \sin^2 20 + \sin^2 30 + \ldots + \sin^2 80\).

1. We can pair the terms such that the sum of angles in each pair is \(90^\circ\):

\(\sin^2 10 + \sin^2 80, \sin^2 20 + \sin^2 70, \sin^2 30 + \sin^2 60, \sin^2 40 + \sin^2 50\)

2. Using the identity \(\sin^2 x + \sin^2 (90 – x) = 1\):

\(\sin^2 10 + \sin^2 80 = 1\)

\(\sin^2 20 + \sin^2 70 = 1\)

\(\sin^2 30 + \sin^2 60 = 1\)

\(\sin^2 40 + \sin^2 50 = 1\)

3. Adding these equations:

\(\sin^2 10 + \sin^2 20 + \sin^2 30 + \sin^2 40 + \sin^2 50 + \sin^2 60 + \sin^2 70 + \sin^2 80 = 4\)

Conclusion:
The value of \(\sin^2 10 + \sin^2 20 + \sin^2 30 + \ldots + \sin^2 80\) is 4.

Given:
– The equation \(\frac{x^{12} + x^3}{x^6} = 0\).

1. Simplify the equation by dividing each term by \(x^6\):
\[ \frac{x^{12}}{x^6} + \frac{x^3}{x^6} = 0 \]
\[ x^6 + \frac{1}{x^3} = 0 \]

2. Rearrange the equation to express \(x^6\) in terms of \(x^3\):
\[ x^6 = -\frac{1}{x^3} \]

3. Raise both sides of the equation to the 3rd power to eliminate the fraction:
\[ (x^6)^3 = \left(-\frac{1}{x^3}\right)^3 \]
\[ x^{18} = -1 \]

4. Further, raise both sides to the 2nd power to find \(x^{36}\):
\[ (x^{18})^2 = (-1)^2 \]
\[ x^{36} = 1 \]

Conclusion:
– The value of \(x^{36}\) is 1.
– Therefore, \(x^{36} + \frac{1}{x^{36}} = 1 + 1 = 2\).

Given:
– The ratio of the curved surface area (CSA) to the total surface area (TSA) of a cylinder is 1:3.
– The total surface area (TSA) is 616 cm².

1. The formula for the TSA of a cylinder is \(2\pi rh + 2\pi r^2\), and the formula for the CSA is \(2\pi rh\).

Given the ratio:
\[ \frac{2\pi rh}{2\pi rh + 2\pi r^2} = \frac{1}{3} \]

2. Solving this equation for \(h\):
\[ 4\pi rh = 2\pi r^2 \]
\[ 2h = r \] (Equation A)

3. Using the given TSA (616 cm²):
\[ 2\pi rh + 2\pi r^2 = 616 \] (Equation B)

4. From equations A and B:
\[ \pi r^2 + 2\pi r^2 = 616 \]
\[ 3\pi r^2 = 616 \]
\[ r^2 = \frac{616 \times 7}{22 \times 3} \]
\[ r^2 = 28 \times \frac{7}{3} \]
\[ r = \frac{14}{\sqrt{3}} \text{ cm} \]

5. Using equation A to find \(h\):
\[ h = \frac{7}{\sqrt{3}} \text{ cm} \]

6. The volume of the cylinder is:
\[ V = \pi r^2 h \]
\[ V = \frac{22}{7} \times \frac{14}{\sqrt{3}} \times \frac{14}{\sqrt{3}} \times \frac{7}{\sqrt{3}} \]
\[ V = \frac{4312}{3\sqrt{3}} \text{ cm}^3 \]

Conclusion:
The volume of water the cylinder can store is \(\frac{4312}{3\sqrt{3}} \text{ cm}^3\).

Let’s denote the distance between the student’s house and the school as \(d\) km.

When the student rides at 5 km/hr, he is 3 minutes late. When he rides at 7 km/hr, he is 3 minutes early. Let’s denote the time it takes to reach the school on time as \(t\) hours.

We can set up two equations based on the information given:

1. When riding at 5 km/hr and being 3 minutes late:

\[ \frac{d}{5} = t + \frac{3}{60} \]

2. When riding at 7 km/hr and being 3 minutes early:

\[ \frac{d}{7} = t – \frac{3}{60} \]

We can solve these two equations simultaneously to find \(d\) and \(t\).

From equation 1:

\[ 60d = 300t + 15 \] (1)

From equation 2:

\[ 60d = 420t – 21 \] (2)

Subtracting equation (2) from equation (1):

\[ 0 = -120t + 36 \]

\[ 120t = 36 \]

\[ t = \frac{36}{120} = \frac{3}{10} \text{ hours} \]

Substituting \(t\) back into either equation (1) or (2) to find \(d\):

\[ 60d = 300 \times \frac{3}{10} + 15 \]

\[ 60d = 90 + 15 \]

\[ 60d = 105 \]

\[ d = \frac{105}{60} = 1.75 \text{ km} \]

Therefore, the distance between the student’s house and the school is 1.75 km.

Let’s denote the simple interest as \(SI\), the principal as \(P\), the rate of interest as \(R\) (in % per annum), and the time of investment as \(T\) (in years).

According to the given information, the ratio of simple interest to principal is \(8 : \frac{25}{2}\), which can be simplified to \(16 : 25\). So, we can write:

\[ \frac{SI}{P} = \frac{16}{25} \]

We also know that the rate of interest is equal to the time invested, so \(R = T\).

The formula for simple interest is:

\[ SI = \frac{P \times R \times T}{100} \]

Substituting \(R = T\) and rearranging the formula, we get:

\[ T^2 = \frac{100 \times SI}{P} \]

Using the ratio \(\frac{SI}{P} = \frac{16}{25}\), we can substitute for \(\frac{SI}{P}\) in the equation:

\[ T^2 = \frac{100 \times 16}{25} \]

Simplifying:

\[ T^2 = \frac{1600}{25} \]

\[ T^2 = 64 \]

Taking the square root of both sides:

\[ T = 8 \text{ years} \]

Therefore, the time of investment is 8 years.