Calculate (1)/(4)A.

$$A = \begin{pmatrix} 4 & -1 \\ 0 & 3 \end{pmatrix}$$ $$\frac{1}{4}A = \begin{pmatrix} \frac{1}{4} \times 4 & \frac{1}{4} \times (-1) \\ \frac{1}{4} \times 0 & \frac{1}{4} \times 3 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{1}{4} \\ 0 & \frac{3}{4} \end{pmatrix}$$

$$B = \begin{pmatrix} -2 & 1 \\ 1 & 0 \end{pmatrix}$$ $$2B = \begin{pmatrix} 2 \times (-2) & 2 \times 1 \\ 2 \times 1 & 2 \times 0 \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 2 & 0 \end{pmatrix}$$

$$\frac{1}{4}A - 2B = \begin{pmatrix} 1 & -\frac{1}{4} \\ 0 & \frac{3}{4} \end{pmatrix} - \begin{pmatrix} -4 & 2 \\ 2 & 0 \end{pmatrix}$$ $$= \begin{pmatrix} 1 - (-4) & -\frac{1}{4} - 2 \\ 0 - 2 & \frac{3}{4} - 0 \end{pmatrix}$$ $$= \begin{pmatrix} 1 + 4 & -\frac{1}{4} - \frac{8}{4} \\ -2 & \frac{3}{4} \end{pmatrix}$$ $$= \begin{pmatrix} 5 & -\frac{9}{4} \\ -2 & \frac{3}{4} \end{pmatrix}$$

The result is $\boxed{\begin{pmatrix} 5 & -\frac{9}{4} \\ -2 & \frac{3}{4} \end{pmatrix}}$.

$$A = \begin{pmatrix} 4 & -1 \\ 0 & 3 \end{pmatrix}$$ $$3A = \begin{pmatrix} 3 \times 4 & 3 \times (-1) \\ 3 \times 0 & 3 \times 3 \end{pmatrix} = \begin{pmatrix} 12 & -3 \\ 0 & 9 \end{pmatrix}$$

$$B + 3A = \begin{pmatrix} -2 & 1 \\ 1 & 0 \end{pmatrix} + \begin{pmatrix} 12 & -3 \\ 0 & 9 \end{pmatrix}$$ $$= \begin{pmatrix} -2 + 12 & 1 + (-3) \\ 1 + 0 & 0 + 9 \end{pmatrix}$$ $$= \begin{pmatrix} 10 & -2 \\ 1 & 9 \end{pmatrix}$$

The result is $\boxed{\begin{pmatrix} 10 & -2 \\ 1 & 9 \end{pmatrix}}$.