Trigonometry: Solving the right angled triangle. SOHCAHTOA. Solve problems 4, 5 and 6.

We have a right-angled triangle with an angle of $41^\circ$. The side adjacent to this angle is 9, and the hypotenuse is 't'.

Since we have the adjacent side and the hypotenuse, we use the cosine function: $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$.

The value of t is $\boxed{11.93}$.

For triangle 5 (finding x):
This triangle is labeled with 'x' for both the hypotenuse and the angle. As presented, with only one side (20) given and two unknowns labeled 'x' (one for the angle and one for the hypotenuse), this problem cannot be solved using standard trigonometric methods. To find 'x' (the side) or 'x' (the angle), either the angle or another side would need to be provided. If the problem implies that the numerical value of the angle is the same as the numerical value of the hypotenuse, it leads to a transcendental equation ($x \sin(x^\circ) = 20$) which is not solvable with basic trigonometry. Therefore, this problem is unsolvable as currently presented due to insufficient information or ambiguous labeling.

For triangle 6 (finding x):

We have a right-angled triangle with an angle of $25^\circ$. The side opposite to this angle is 'x', and the side adjacent to this angle is 12.

Since we have the opposite side and the adjacent side, we use the tangent function: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$.

The value of x is $\boxed{5.59}$.