Let’s first find the work done by each group in one day, which we can call their work rate. We’ll denote the total work to develop the app as 1 unit of work.

– 4 men can develop the app in 3 days, so their work rate is \( \frac{1}{3 \times 4} = \frac{1}{12} \) of the work per day per man.
– 3 women can develop the app in 6 days, so their work rate is \( \frac{1}{6 \times 3} = \frac{1}{18} \) of the work per day per woman.
– 6 boys can develop the app in 4 days, so their work rate is \( \frac{1}{4 \times 6} = \frac{1}{24} \) of the work per day per boy.

Now, 3 men and 6 boys worked together for 1 day. The work done by them in 1 day is:

\[ 3 \times \frac{1}{12} + 6 \times \frac{1}{24} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \text{ of the work} \]

So, half of the work is remaining.

If only women are to finish the remaining half of the work in 1 day, the number of women required is:

\[ \text{Number of women} = \frac{\text{Remaining work}}{\text{Work rate of one woman}} = \frac{1/2}{1/18} = 9 \text{ women} \]

Therefore, 9 women would be required to finish the remaining work in 1 day.

Let’s denote the weight of water in the fresh fruits as \(W_f\) kg and the total weight of the fresh fruits as \(F\) kg.

According to the given information, the ratio of the quantity of water in fresh fruits to that of dry fruits is 7:2. This means that for every 7 kg of water in fresh fruits, there are 2 kg of water in dry fruits.

We know that 400 kg of dry fruits contain 50 kg of water. Using the given ratio, we can find the weight of water in the fresh fruits:

\[ \frac{W_f}{50 \text{ kg}} = \frac{7}{2} \]

Solving for \(W_f\):

\[ W_f = 50 \text{ kg} \times \frac{7}{2} = 175 \text{ kg} \]

Therefore, the weight of the water in the same fruits when they were fresh is 175 kg.

We can rewrite the equation \(2^{2x – 1} = \frac{1}{8^{x – 3}}\) using the properties of exponents:

\[2^{2x – 1} = 2^{-3(x – 3)}\]

This is because \(8 = 2^3\), so \(8^{x – 3} = (2^3)^{x – 3} = 2^{3(x – 3)}\).

Since the bases are the same, we can set the exponents equal to each other:

\[2x – 1 = -3(x – 3)\]

Expanding:

\[2x – 1 = -3x + 9\]

Adding \(3x\) to both sides and adding 1 to both sides:

\[5x = 10\]

Dividing both sides by 5:

\[x = 2\]

Therefore, the value of \(x\) is 2.

To find the height of the balloon from the ground, we can use trigonometry. We know that the cable is inclined at a 60° angle to the horizontal and has a length of 200 m.

The height of the balloon (opposite side) can be found using the sine function:

\[
\sin(\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}
\]

In this case:

\[
\sin(60°) = \frac{\text{height}}{200 \text{ m}}
\]

Solving for the height:

\[
\text{height} = 200 \text{ m} \times \sin(60°)
\]

The value of \(\sin(60°)\) is \(\frac{\sqrt{3}}{2}\), so:

\[
\text{height} = 200 \text{ m} \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \text{ m}
\]

Therefore, the height of the balloon from the ground is \(100\sqrt{3}\) meters.

Given:

\[
\cot \left(\frac{\pi}{2} – \frac{\theta}{2}\right) = \sqrt{3}
\]

We can use the cofunction identity \(\cot(\frac{\pi}{2} – x) = \tan(x)\) to rewrite the given equation:

\[
\tan\left(\frac{\theta}{2}\right) = \sqrt{3}
\]

Now, we know that \(\tan(\frac{\pi}{3}) = \sqrt{3}\), so:

\[
\frac{\theta}{2} = \frac{\pi}{3}
\]

Thus:

\[
\theta = \frac{2\pi}{3}
\]

Now, we can find the value of \(\sin \theta – \cos \theta\):

\[
\sin \theta – \cos \theta = \sin\left(\frac{2\pi}{3}\right) – \cos\left(\frac{2\pi}{3}\right)
\]

Using the values of sine and cosine for \(\frac{2\pi}{3}\):

\[
\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}
\]

So:

\[
\sin \theta – \cos \theta = \frac{\sqrt{3}}{2} – \left(-\frac{1}{2}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}
\]

Therefore, the value of \(\sin \theta – \cos \theta\) is \(\frac{\sqrt{3} + 1}{2}\).