34 + 36 + 38 + 40 +42 + ... + 88 + 90

Let's solve each of the problems you've listed step by step.

### Problem 1:
**34 + 36 + 38 + 40 + 42 + ... + 88 + 90**

This is an arithmetic series where:
- First term \(a = 34\)
- Common difference \(d = 2\)
- Last term \(l = 90\)

To find the number of terms \(n\):
\[ l = a + (n-1)d \]
\[ 90 = 34 + (n-1) \cdot 2 \]
\[ 90 - 34 = (n-1) \cdot 2 \]
\[ 56 = (n-1) \cdot 2 \]
\[ n-1 = 28 \]
\[ n = 29 \]

Now find the sum \(S_n\):
\[ S_n = \frac{n}{2} (a + l) \]
\[ S_{29} = \frac{29}{2} (34 + 90) \]
\[ S_{29} = \frac{29}{2} \cdot 124 \]
\[ S_{29} = 29 \cdot 62 \]
\[ S_{29} = 1798 \]

### Problem 2:
**34 + 36 + 38 + 40 + 42 + ... + 88 + 90** (This is the same as Problem 1 and we already calculated it.)

**Answer: 1798**

### Problem 3:
**35 * 34 + 35 * 86 + 65 * 75 + 65 * 45**

Breaking it down:
\[ 35 \cdot 34 = 1190 \]
\[ 35 \cdot 86 = 3010 \]
\[ 65 \cdot 75 = 4875 \]
\[ 65 \cdot 45 = 2925 \]

Now, sum them up:
\[ 1190 + 3010 + 4875 + 2925 \]
\[ = 1190 + 3010 = 4200 \]
\[ = 4200 + 4875 = 9075 \]
\[ = 9075 + 2925 = 12000 \]

### Problem 4:
**789 + 100 : [ 16 - 2 (5² - 22)]**

Calculating inside the brackets first:
\[ 5² = 25 \]
\[ 5² - 22 = 25 - 22 = 3 \]
\[ 2(3) = 6 \]
\[ 16 - 6 = 10 \]

Now:
\[ 100 : 10 = 10 \]

Finally:
\[ 789 + 10 = 799 \]

### Problem 5:
**10² - [ 160 : (5⁶ - 3 \cdot 5)]²**

Calculating \(10²\):
\[ 10² = 100 \]

Calculating \(5⁶\) and \(3 \cdot 5\):
\[ 5⁶ = 15625 \]
\[ 3 \cdot 5 = 15 \]
\[ 5⁶ - 15 = 15625 - 15 = 15610 \]
\[ 160 : 15610 \approx 0.01024 \]

Now calculating the square:
\[ (0.01024)² = 0.0001048576 \]

So finally:
\[ 100 - 0.0001048576 \approx 99.9998951424 \]

**Final Answers:**
1) 1798
2) 1798
3) 12000
4) 799
5) Approximately 99.9999