a.
The answer is $a_n = 2 \cdot 3^n - 2^n$.
The characteristic equation is $\lambda^2 - 5\lambda + 6 = 0$. The roots are $\lambda = 3, 2$. Therefore, we have
$$a_n = A \cdot 3^n + B \cdot 2^n$$
for some constants $A$ and $B$. Putting $n = 0$ and $n = 1$, we need to solve
$$\begin{cases} A + B = 1, \\ 3A + 2B = 4. \end{cases}$$
We easily deduce $A = (3A + 2B) - 2(A + B) = 2$ and $B = 1 - A = -1$. Thus, we have
$$a_n = 2 \cdot 3^n - 2^n.$$

b.
The answer is $b_n = (n + 6)2^{n-1}$.
The characteristic equation is $\lambda^2 - 4\lambda + 4 = 0$, with a double root $\lambda = 2$. Therefore, we have
$$b_n = (Cn + D)2^n$$
for some constants $C$ and $D$. Putting $n = 0$ and $n = 1$, we need to solve
$$\begin{cases} D = 3, \\ 2(C + D) = 7. \end{cases}$$
Clearly, $C = \frac{7}{2} - D = \frac{1}{2}$. Thus, we have
$$b_n = (n + 6)2^{n-1}.$$