We substitute $x_n = 4y_n + 2$. Then the equation becomes homogeneous:
$$x_{n+1} + x_{n-1} = 4y_{n+1} + 2 + 4y_{n-1} + 2 = 4(4y_n + 1) + 4 = 16y_n + 8 = 4x_n,$$
with initial conditions $x_0 = 4(-\frac{1}{4}) + 2 = 1$ and $x_1 = 4 \cdot 0 + 2 = 2$. So all the numbers in the sequence $(x_i)$ are integers and we see that $x_{n+1}$ and $x_{n-1}$ always have the same parity. In particular, $x_{2n}$ is always odd. So $2y_{2n} + \frac{3}{2} = \frac{4y_{2n}+3}{2} = \frac{x_{2n+1}}{2}$ is always an integer. To show that they are also positive, we prove by induction that $x_n$ is an increasing sequence of positive numbers. This is indeed true for $x_1 > x_0 > 0$. Now suppose as induction hypothesis that $x_n > x_{n-1} > 0$. Then we also find that $x_{n+1} - x_n = 3x_n - x_{n-1} > x_n - x_{n-1} > 0$. With this, we conclude the proof of part (a).

For part (b), we note that the characteristic equation for the homogeneous part $y_{n+1} + y_{n-1} = 4y_n$ is given by $x^2 + 1 = 4x$. Of these, the solutions are $x = 2 - \sqrt{3}$ and $x = 2 + \sqrt{3}$. Now we choose a solution to the inhomogeneous equation, say $y_n = -\frac{1}{2}$. Then the general solution is
$$y_n = A(2 - \sqrt{3})^n + B(2 + \sqrt{3})^n - \frac{1}{2}.$$
If we solve this using $n = 0$ and $n = 1$, then we find $-\frac{1}{4} = A + B - \frac{1}{2}$ and
$0 = A(2 - \sqrt{3}) + B(2 + \sqrt{3}) - \frac{1}{2} = 2(A + B) + \sqrt{3}(B - A) - \frac{1}{2}$. This means
that $A + B = \frac{1}{4}$ and $B - A = 0$, or $A = B = \frac{1}{8}$. So
$$y_n = \frac{1}{8}(2 - \sqrt{3})^n + \frac{1}{8}(2 + \sqrt{3})^n - \frac{1}{2}.$$

$$\begin{align*} (4y_n + 2)^2 &= \left(\frac{1}{2}(2 - \sqrt{3})^n + \frac{1}{2}(2 + \sqrt{3})^n\right)^2 \\ &= \frac{1}{4}(2 - \sqrt{3})^{2n} + \frac{1}{4}(2 + \sqrt{3})^{2n} + \frac{1}{2}(2 - \sqrt{3})^n(2 + \sqrt{3})^n \\ &= \frac{1}{4}(2 - \sqrt{3})^{2n} + \frac{1}{4}(2 + \sqrt{3})^{2n} + \frac{1}{2} \\ &= 2y_{2n} + \frac{3}{2}. \end{align*}$$
This proves part (b), because $4y_n + 2 = x_n$ is an integer. $\square$