This is the impulse response of a discrete system. X1 , X2, X3,X4 were unknowns and that's why the picture might not seem right. X1=1, X2=5, X3=2, X4=7
I might have done this right but I'm having second thoughts since I haven't practised much on these.$$T(k)=\frac{Y(k)}{δ(k)}$$ where δ(k) is the input Y(k) the output and T(k) my transfer function$$Y(k)=δ(k)T(k)\\Y(k)=δ(k)(5u(k-1)-3u(k-2)+5(k-3)-7u(k-4)$$Z-transforming I get: $$G(z)=\frac{5z}{z-1}z^{-1}-\frac{3z}{z-1}z^{-2}+\frac{5z}{z-1}z^{-3}-\frac{7z}{z-1}z^{-4}$$We can go on and simplify ... It seemed too easy so if anyone could confirm I'd appreciate it.
I guess you are dealing with a finite impulse response consisting of $4$ taps (non-zero elements). In that case the impulse response can be described (in time domain) as an impulse train
$$ h[n] = \sum_{k=1}^4X_k \delta[n-kn_k],n\in \mathbb{Z}, $$
where $n_k$ is the delay (shift) of the $k$-th tap. Now you can find the transfer function $\mathcal{Z}(h[n])$ by using $\mathcal{Z}(\delta[n-l])=z^{-l},l\in\mathbb{Z}$, and the linearity of the $\mathcal{Z}$ transform.
P.S.: Yes, this is an easy exercise, however, your solution is wrong and your notation is a bit sloppy.