1. Let $a_{n}$ be the no. of ternary strings not containing "00" or "11" of length $n$. Let $a_{0_{n}}$ be the no. of ternary strings not containing "00" or "11" of length $n$ and ending in '0' and similarly we define $a_{1_n}$ and $a_{2_{n}}$. So,
$$\begin{align*} a_{n} &= a_{0_{n}} + a_{1_n} + a_{2_{n}} \\
&= \left(a_{1_{n-1}} + a_{2_{n-1}}\right) + \left(a_{0_{n-1}}+a_{2_{n-1}}\right) + \left(a_{0_{n-1}} + a_{1_{n-1}} + a_{2_{n-1}}\right)\\
&= 2. \left(a_{0_{n-1}} + a_{1_{n-1}} + a_{2_{n-1}}\right) + a_{2_{n-1}}\\
&= 2.a_{n-1} + a_{n-2}. \end{align*}$$
For initial condition, $a_1 = 3, a_2 = 7.$(All two length strings except 11 and 00).
2. We can get this as
$$b_n = 3^n - a^n.$$
where $a_n$ is as in 1. Solving, we get
$$\begin{align*}b_n &= 3^n - 2.a_{n-1} - a_{n-2} \\
&= 3^n - 2. \left(3^{n-1} - b^{n-1}\right) - 3^{n-2} + b^{n-2} \\
&= 3^{n-2} \left[ 3^2 - 2.3 - 1 \right] + 2.b^{n-1} + b^{n-2} \\
&= 2. \left(3^{n-2} + b^{n-1} \right) + b^{n-2} \end{align*}$$
Initial condition, $b_1 = 0, b_2 = 2.$
3. Similar to 1, we can write
$$\begin{align*} a_n &= a0_n + a_{1_n} + a_{2_{n}} \\
&= a_{1_{n-1}} + a_{2_{n-1}} + a_{0_{n-1}} + a_{2_{n-1}} + a_{0_{n-1}} + a_{1_{n-1}}\\
&= 2. a_{n-1}. \end{align*} $$
Initial condition $a_1 = 3, a_2 = 6.$
4. $$b_n = 3^n - a_n$$
where $a_n = 2.a_{n-1}$ as shown in 3. Solving we get,
$$\begin{align*} b^n &= 3^n - 2. a_{n-1} \\
&= 3^n - 2. \left(3^{n-1} - b_{n-1}\right). \end{align*}$$
Initial condition $b_1 = 0, b_2 = 3.$
