To solve the given systems of equations, let's break them down one by one.

### 1) System of equations:

\[
\begin{cases}
\frac{1}{x+1} + \frac{1}{y-2} = 3 \\\\
\frac{2}{x+1} + \frac{1}{y-2} = 1
\end{cases}
\]

Let's denote \( a = \frac{1}{x+1} \) and \( b = \frac{1}{y-2} \).

This converts the system into:

\[
\begin{cases}
a + b = 3 \\\\
2a + b = 1
\end{cases}
\]

Subtract the first equation from the second:

\[
(2a + b) - (a + b) = 1 - 3 \\
a = -2
\]

Substituting \( a \) back into the first equation:

\[
-2 + b = 3 \\
b = 5
\]

Now substitute back to find \( x \) and \( y \):

\[
\frac{1}{x+1} = -2 \implies x + 1 = -\frac{1}{2} \implies x = -\frac{3}{2} \\
\frac{1}{y-2} = 5 \implies y - 2 = \frac{1}{5} \implies y = 2 + \frac{1}{5} = \frac{11}{5}
\]

---

### 2) System of equations:

\[
\begin{cases}
\frac{2}{y-3} = -9 \\\\
\frac{1}{x-1} = 7 \\\\
\frac{1}{y-3} = -7
\end{cases}
\]

From the first:

\[
\frac{2}{y-3} = -9 \implies y - 3 = -\frac{2}{9} \implies y = 3 - \frac{2}{9} = \frac{25}{9}
\]

From the second:

\[
\frac{1}{x-1} = 7 \implies x - 1 = \frac{1}{7} \implies x = 1 + \frac{1}{7} = \frac{8}{7}
\]

Now check the third equation:

\[
\frac{1}{y-3} = -7 \implies y - 3 = -\frac{1}{7} \implies y = 3 - \frac{1}{7} = \frac{20}{7}
\]
This shows inconsistency, likely needing checking the premises.

---

### 3) Calculation:

\[
\begin{cases}
\frac{1}{x+1} + \frac{4}{y+2} = 2 \\
\frac{2}{x+1} + \frac{2}{y+2} = 1
\end{cases}
\]

Substituting similar variables as before:

1. Let \( a = \frac{1}{x+1} \) and \( b = \frac{1}{y+2} \).

The equations become:

\[
\begin{cases}
a + 4b = 2 \\
2a + 2b = 1
\end{cases}
\]

From the second equation:

\[
a + b = \frac{1}{2} \implies b = \frac{1}{2} - a
\]

Substituting into the first equation:

\[
a + 4\left(\frac{1}{2} - a\right) = 2 \\
a + 2 - 4a = 2 \\
-3a = 0 \implies a = 0 \implies x+1 = \infty \implies x = -1 \\
b = \frac{1}{2} \implies y + 2 = 2 \implies y = 0
\]

### Final Summary:

1. \( x = -\frac{3}{2}, y = \frac{11}{5} \)
2. Inconsistencies; check equations for errors; may need to confirm setup.
3. \( x = -1, y = 0 \) for \( \frac{1}{x+1}, \frac{1}{y+2} \).

Please review or let me know for further specifics!