The image contains four geometry problems involving right triangles with labeled sides and angles.

This is a right-angled triangle. The hypotenuse is 10, one leg is 7, and the other leg is $e$.

The Pythagorean theorem states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs and $c$ is the hypotenuse.

The value of $e$ is $\boxed{\sqrt{51}}$.

2) Find the values of $s$ and $a$.

This is a right-angled triangle. One leg is 6, the hypotenuse is $s$, and the angle opposite the leg of length 6 is $a$.

Using the sine function, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$.

With only one side and one angle (both unknown) related by this equation, it is not possible to find unique numerical values for both $s$ and $a$ without additional information (e.g., another side length or another angle). The relationship between them is $s = \frac{6}{\sin(a)}$ or $a = \arcsin\left(\frac{6}{s}\right)$.

3) Find the length of the unlabeled side.

This is a right-angled triangle. The legs are 7 and 9. The hypotenuse is the unlabeled side. Let's call it $h$.

The length of the unlabeled side is $\boxed{\sqrt{130}}$.

4) Find the value of $x$.

This is a right-angled triangle. The hypotenuse is 7, one leg is 3, and the other leg is $x$. An angle of $37^\circ$ is also given.

From the perspective of the $37^\circ$ angle, the adjacent side is 3 and the hypotenuse is 7.
If these values were consistent, $\cos(37^\circ)$ should be equal to $\frac{3}{7}$.
$\cos(37^\circ) \approx 0.7986$.
$\frac{3}{7} \approx 0.4286$.
Since $0.7986 \neq 0.4286$, the given angle of $37^\circ$ is inconsistent with the given side lengths of 3 and 7.

The value of $x$ is $\boxed{2\sqrt{10}}$.