What are the solutions to these polynomial division problems?

The divisor is $(x-7)$, so we use $k=7$. The coefficients of the dividend are $1, -5, 1, 42$.

$$\begin{array}{c|cccc} 7 & 1 & -5 & 1 & 42 \\ & & 7 & 14 & 105 \\ \hline & 1 & 2 & 15 & 147 \\ \end{array}$$

The last number, 147, is the remainder. The other numbers, $1, 2, 15$, are the coefficients of the quotient, starting with $x^2$.

The quotient is $x^2 + 2x + 15$ with a remainder of 147.
So, $\frac{x^3 - 5x^2 + x + 42}{x-7} = x^2 + 2x + 15 + \frac{147}{x-7}$.

The result of the division is $\boxed{x^2 + 2x + 15 \text{ with a remainder of } 147}$.

2. Divide $2x^3 - 15x^2 - 2x + 120$ by $(x-6)$ and $(x-4)$

This means dividing by $(x-6)(x-4)$. We can perform two successive synthetic divisions.

Use $k=6$.

$$\begin{array}{c|cccc} 6 & 2 & -15 & -2 & 120 \\ & & 12 & -18 & -120 \\ \hline & 2 & -3 & -20 & 0 \\ \end{array}$$

The remainder is 0. The quotient is $2x^2 - 3x - 20$.

Use $k=4$.

$$\begin{array}{c|ccc} 4 & 2 & -3 & -20 \\ & & 8 & 20 \\ \hline & 2 & 5 & 0 \\ \end{array}$$

The remainder is 0. The final quotient is $2x + 5$.

The result of the division is $\boxed{2x + 5}$.

3. Divide $6x^4 + 13x^3 - 36x^2 - 43x + 30$ by $(x-2)$

The divisor is $(x-2)$, so we use $k=2$. The coefficients of the dividend are $6, 13, -36, -43, 30$.

$$\begin{array}{c|ccccc} 2 & 6 & 13 & -36 & -43 & 30 \\ & & 12 & 50 & 28 & -30 \\ \hline & 6 & 25 & 14 & -15 & 0 \\ \end{array}$$

The last number, 0, is the remainder. The other numbers, $6, 25, 14, -15$, are the coefficients of the quotient, starting with $x^3$.

The quotient is $6x^3 + 25x^2 + 14x - 15$ with a remainder of 0.
So, $\frac{6x^4 + 13x^3 - 36x^2 - 43x + 30}{x-2} = 6x^3 + 25x^2 + 14x - 15$.

The result of the division is $\boxed{6x^3 + 25x^2 + 14x - 15}$.