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 \(\frac{\sqrt{3} + 1}{2}\).

To find the values of \(\sin x\) given the equation \(8\sin x = 4 + \cos x\), we can use trigonometric identities to rewrite the equation in terms of a single trigonometric function.

The given equation is:
\[8\sin x = 4 + \cos x\]

We know the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\). Our goal is to express \(\cos x\) in terms of \(\sin x\) (or vice versa) to solve for \(\sin x\). Since the equation involves both \(\sin x\) and \(\cos x\), and we’re looking to find \(\sin x\), let’s isolate \(\cos x\) and then use the Pythagorean identity.

Rearrange the given equation to isolate \(\cos x\):
\[\cos x = 8\sin x – 4\]

Using the Pythagorean identity \(\cos^2 x = 1 – \sin^2 x\), we substitute for \(\cos x\) in the equation:
\[1 – \sin^2 x = (8\sin x – 4)^2\]

Expanding the right side and moving all terms to one side gives us a quadratic equation in \(\sin x\):
\[1 – \sin^2 x = 64\sin^2 x – 64\sin x + 16\]

Combine like terms:
\[65\sin^2 x – 64\sin x + 15 = 0\]

This is a quadratic equation in \(\sin x\). To solve for \(\sin x\), we use the quadratic formula where \(a = 65\), \(b = -64\), and \(c = 15\):
\[\sin x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]
\[= \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 65 \cdot 15}}{2 \cdot 65}\]
\[= \frac{64 \pm \sqrt{4096 – 3900}}{130}\]
\[= \frac{64 \pm \sqrt{196}}{130}\]
\[= \frac{64 \pm 14}{130}\]

So, we have two possible solutions for \(\sin x\):
1. \(\sin x = \frac{64 + 14}{130} = \frac{78}{130} = \frac{39}{65} = \frac{3}{5}\)
2. \(\sin x = \frac{64 – 14}{130} = \frac{50}{130} = \frac{25}{65} = \frac{5}{13}\)

Therefore, the values of \(\sin x\) that satisfy the given equation are \(\frac{3}{5}\) and \(\frac{5}{13}\).