角 $A$ 和 $B$ 均为钝角，且 $\tan(A + B) = \frac{3}{4}$。 在不使用计算器的前提下：

1. Determine the initial possible range for $(A + B)$ given that $A$ and $B$ are obtuse

Since $A$ and $B$ are obtuse,
$90^\circ < A < 180^\circ$ and $90^\circ < B < 180^\circ$
Therefore, $180^\circ < (A + B) < 360^\circ$

2. Use the sign of the tangent function to narrow down the quadrant

We are given $\tan(A + B) = \frac{3}{4} > 0$.
Since tangent is positive in the 1st and 3rd quadrants, and $180^\circ < (A + B) < 360^\circ$, $(A + B)$ must lie in the 3rd quadrant.
Thus, $180^\circ < (A + B) < 270^\circ$.

1. Determine the quadrant of $(A + B)$

Since $A$ and $B$ are obtuse, $90^\circ < A < 180^\circ$ and $90^\circ < B < 180^\circ$, so $180^\circ < A + B < 360^\circ$.
Given $\tan(A + B) = \frac{3}{4} > 0$, $(A + B)$ must lie in the third quadrant ($180^\circ < A + B < 270^\circ$).

2. Find the value of $\sin(A + B)$

In the third quadrant, sine is negative.
Using a right-angled triangle with opposite side $3$ and adjacent side $4$, the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Therefore, $\sin(A + B) = -\frac{3}{5}$.

1. Set up the equation for $\tan A$ using the tangent addition formula.

We know that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Substituting the given values $\tan(A + B) = \frac{3}{4}$ and $\tan B = -\frac{12}{5}$:
$\frac{3}{4} = \frac{\tan A - \frac{12}{5}}{1 - \tan A \left(-\frac{12}{5}\right)}$

2. Solve the equation for $\tan A$.

$\frac{3}{4} = \frac{\tan A - 2.4}{1 + 2.4\tan A}$
$3(1 + 2.4\tan A) = 4(\tan A - 2.4)$
$3 + 7.2\tan A = 4\tan A - 9.6$
$3.2\tan A = -12.6$
$\tan A = -\frac{12.6}{3.2} = -\frac{126}{32} = -\frac{63}{16}$

3. Find $\cos A$ from $\tan A$, considering the quadrant of angle $A$.

Since $A$ is an obtuse angle, it lies in the second quadrant, where cosine is negative.
Using a right-angled triangle with opposite side $63$ and adjacent side $16$:
$\text{Hypotenuse} = \sqrt{63^2 + 16^2} = \sqrt{3969 + 256} = \sqrt{4225} = 65$
Therefore, $\cos A = -\frac{\text{adjacent}}{\text{hypotenuse}} = -\frac{16}{65}$.