Answer: 200 N and 1.572 m/s$^2$

Let us draw the Free body diagram for the problem stated.

Applying Newton's Second Law separately to the i and j components:

Along the y-axis, $F_y = ma_y \rightarrow N - m_1g - F_1 \sin 35 + F_2 \sin 43 = 0$

$\Rightarrow N = 20 \times 9.8 + 25 \sin 35 - 15 \sin 43 = 200.3\; N$

Along the x-axis, $F_x = ma_x \rightarrow F_1 \cos 35 + F_2 \cos 43 =m_1 a$

$\Rightarrow a = \large\frac{25 \cos 35 + 15 \cos 43}{20}$$ = 1.572\;m/s^2$