I'm partial to solving with generating functions. Let
[tex]T(x)=\displaystyle\sum_{n\ge0}t_nx^n[/tex]
Multiply both sides of the recurrence by [tex]x^{n+2}[/tex] and sum over all [tex]n\ge0[/tex].
[tex]\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}[/tex]
Shift the indices and factor out powers of [tex]x[/tex] as needed so that each series starts at the same index and power of [tex]x[/tex].
[tex]\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n[/tex]
Now we can write each series in terms of the generating function [tex]T(x)[/tex]. Pull out the first few terms so that each series starts at the same index [tex]n=0[/tex].
[tex]2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)[/tex]
Solve for [tex]T(x)[/tex]:
[tex]T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}[/tex]
Splitting into partial fractions gives
[tex]T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}[/tex]
which we can write as geometric series,
[tex]T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n[/tex]
[tex]T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n[/tex]
which tells us
[tex]\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}[/tex]
# # #
Just to illustrate another method you could consider, you can write the second recurrence in matrix form as
[tex]49y_{n+2}=-16y_n\implies y_{n+2}=-\dfrac{16}{49}y_n\implies\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}\begin{bmatrix}y_{n+1}\\y_n\end{bmatrix}[/tex]
By substitution, you can show that
[tex]\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}[/tex]
or
[tex]\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}[/tex]
Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of [tex]n-1[/tex], then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.
