To solve these problems, let's tackle them step by step:

### Bài 1: Tìm \( x \), biết:

#### a)
\[ x^3 + 9x^2 + 27x + 19 = 0 \]

This is a cubic equation. You may need to use either factorization (if possible) or the Rational Root Theorem to find potential solutions.

#### b)
\[ 25(x+3)^2 + (1-5x)(1+5x) = 8 \]

Simplify the equation:

1. Expand and simplify \((1-5x)(1+5x)\) using the difference of squares formula: \(a^2-b^2 = (a-b)(a+b)\).
2. Solve the quadratic equation.

#### c)
\[ 3(x+2)^2 + (2x-1)^2 - 7(x+3)(x-3) = 36 \]

1. Simplify each part: \((x+2)^2\), \((2x-1)^2\), and \((x+3)(x-3)\).
2. Combine all terms and solve the resulting quadratic equation.

### Bài 2: Cho hai hình bình hành \( MNBA \) và \( MNCB \).

#### a)
Chứng minh \( B \) là trung điểm của \( AC \).

In a parallelogram, the diagonals bisect each other. Use this property to prove that \( B \) is the midpoint.

#### b)
Hỏi tam giác \( MAB \) thỏa mãn điều kiện gì để \( MNCA \) là một hình thang cân?

For \( MNCA \) to be an isosceles trapezoid:
- Two sides must be parallel, and the non-parallel sides equal.
- \( MAB \) must satisfy specific parallel and equality conditions.

#### c)
Lấy điểm \( D \) để từ giác \( MNDC \) là hình bình hành. Hỏi tam giác \( MAB \) thỏa mãn điều kiện gì để \( MNDA \) là một hình thang cân?

For \( MNDC \) to be a parallelogram, opposite sides must be equal and parallel.
For \( MNDA \) to be an isosceles trapezoid, similar conditions as part b apply, with proper adjustments based on different vertices.

These problems require an understanding of both algebraic manipulation and geometric properties of shapes. You can solve them by systematically applying these principles.