To solve these problems, we need to find the other trigonometric values given the conditions.

### Problem 1:
Given:
\[
\cos x = \frac{2}{\sqrt{5}}, \quad -\frac{\pi}{2} < x < 0
\]

Since \( x \) is in the fourth quadrant, \(\sin x\) will be negative. We use the Pythagorean identity:

\[
\sin^2 x + \cos^2 x = 1
\]

\[
\sin^2 x + \left(\frac{2}{\sqrt{5}}\right)^2 = 1
\]

\[
\sin^2 x + \frac{4}{5} = 1
\]

\[
\sin^2 x = \frac{1}{5}
\]

\[
\sin x = -\frac{1}{\sqrt{5}}
\]

Now, \(\tan x\) is:

\[
\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = -\frac{1}{2}
\]

Thus, the other trigonometric values are:
\[
\sin x = -\frac{1}{\sqrt{5}}, \quad \tan x = -\frac{1}{2}
\]

### Problem 2:
Given:
\[
\tan x = -\frac{3}{4}, \quad -\pi < x < -\frac{\pi}{2}
\]

Since \( x \) is in the third quadrant, both \(\sin x\) and \(\cos x\) are negative. We use the identity:

\[
\tan x = \frac{\sin x}{\cos x}
\]

Let \(\sin x = -3k\) and \(\cos x = -4k\). Then:

\[
\tan x = \frac{-3k}{-4k} = \frac{3}{4}
\]

Now, using:

\[
\sin^2 x + \cos^2 x = 1
\]

\[
(-3k)^2 + (-4k)^2 = 1
\]

\[
9k^2 + 16k^2 = 1
\]

\[
25k^2 = 1
\]

\[
k^2 = \frac{1}{25}
\]

\[
k = \frac{1}{5}
\]

So:
\[
\sin x = -3k = -\frac{3}{5}, \quad \cos x = -4k = -\frac{4}{5}
\]

Thus, the other trigonometric values are:
\[
\sin x = -\frac{3}{5}, \quad \cos x = -\frac{4}{5}
\]