Output of a discrete-time LTI system different form than input?
NOTE: The answer is not complete; however, it seems the problem cannot be solved.

We have the following z-transforms of the input signals: 
$$x_1[n]=(1/6)^n\cdot u[n]\quad\longleftrightarrow\quad X_1(z)=\frac{1}{1-(1/6)\cdot z^{-1}}$$
$$x_2[n] = (-1)^n\quad\longleftrightarrow\quad X_2(z)=\frac{1}{1+z^{-1}}$$

For the output signals we have the following z-transforms:
$$y_1[n] = [a(1/2)^2 + 10(1/3)^n]u[n] \quad\longleftrightarrow\quad Y_1(z) = a\frac{1}{1-(1/2)\cdot z^{-1}} + 10\frac{1}{1-(1/3)\cdot z^{-1}}$$
$$y_2[n] = (7/4)(-1)^n\quad\longleftrightarrow\quad Y_2(z) = (7/4)\frac{1}{1+z^{-1}}$$

For both pairs of signals we know that the transfer function equals $H(z)=\frac{Y(z)}{X(z)}$. From $\{X_2(z),Y_2(z)\}$ we know that $H(z) = 7/4$ (Which is trivial due to linearity, i.e., it follows also directly from $\{x_2[n],y_2[n]\}$).

What remains is to solve $Y_1(z) = H(z)\cdot X_1(z)$ for $a$; however, I don't get a proper result.

EDIT: Another way to appraoch this last step is to solve the following: 
$$a\frac{2}{2-z^{-1}} + b\frac{3}{3-z^{-1}} = c\frac{6}{6-z^{-1}}$$
where you may choose $b=10$ if desired. 

In matrix representation:
$$\mathbf{A}\left[\begin{array}{1}a\\b\\c\end{array}\right] = \mathbf{b}$$
the matrix $\mathbf{A}$ is a full rank size 3x3 matrix, which shows that no scalar solution for $H(z)=c$ with $b=10$ exists.

EDIT2:
The solution seems to be that $H(-1)=7/4$ and $a=-9$. We already found that $H(z)=7/4$ for any $z$, based on the signals $\{X_2(z), Y_2(z)\}$. When evaluating $\{X_1(z),Y_1(z)\}$ for $z=-1$, the solution $a=-9$ follows easily. However, for another value of $z$, say $z=-2$ we obtain $a\approx -9.375$.

Conclusion, the decomposition only holds for $z=-1$.