Since number $7$ requires a solution to a system of equations I thought some elimination would be appropriate. We know that $Q'(x)$ must be continuous at all of the endpoints for this to be a spline, so let's start there:

First I found the slope at the right endpoint of $\frac{1}{2}x^2+2x+\frac{3}{2}$ which was $lim_{x\rightarrow1^-}Q'(x)$ or $1+2$ or $3$. That means that $2\alpha+\beta = 3$. Then I found the slope at the left endpoint of $3x^2-7x+12$ which was $\lim_{x\rightarrow3^+}Q'(x)$ or $18-7$ or $11$. This gives me $6\alpha+\beta = 11$. Organizing this in a $2$x$3$ matrix I start with:

\[\left [\begin{array}{c} \alpha \\ \beta \end{array} \right] \left[ \begin{array}{cc} 2 & 1 \\ 6 & 1 \end{array} \right] = \left [\begin{array}{c} 3 \\ 11 \end{array} \right]\]

Which I organize into a $2$x$3$ augmented matrix and solve using elimination:

\[ \left [\begin{array}{cc|c}2 & 1 & 3 \\ 6 & 1 & 11\end{array} \right] \rightarrow \left[ \begin{array}{cc|c}2 & 1 & 3 \\ 4 & 0 & 8\end{array} \right] \rightarrow \left[ \begin{array}{cc|c}2 & 1 & 3 \\ 1 & 0 & 2\end{array} \right] \rightarrow \left[ \begin{array}{cc|c}0 & 1 & -1 \\ 1 & 0 & 2\end{array} \right]

\]

As you can see, I got $\alpha = 2$ and $\beta = -1$. From here it's just a matter of plugging $\alpha$ and $\beta$ into $Q(x)$ and solving for $\gamma$ to make sure $Q(x)$ is continuous. Plugging in at the bound where $x = 1$ gives me $2-1+\gamma = 4$ which I easily solved for $\gamma = 3$. Just to be sure, I plugged this into the bound at $x = 3$ to make sure it would match up and got $18-3+3 = 18$, which I believe to be true.

So I found $\alpha = 2$, $\beta = -1$ and $\gamma = 3$ ensure $Q(x)$ is a spline, amidoinitrite?