The problem reads as follows:
A slight generalization of the Taylor expansion is
$$f(x+a)=\sum^\infty_{n=0}\frac{a^n}{n!}f^{(n)}(x)=f(x+a)$$
Calculate the functions $e^{kd/dt}f(t)$ and $e^{td/dt}f(t)$
The answer in each case is a simple, analytic function, not a series.
Im confused as to why he would provide the Taylor series expansion but we will not use it for our answer?
You've managed the first one in the comments, so I'll look at the second one. We have two ways in: we can either write $e^{td/dt}$ as a series and apply it to $f(t)$, or expand $f(t)$ as a series $\sum_{k} f^{(k)}(0)t^k/k!$ and act on the individual terms. Let's try the latter. We need to know $$ e^{td/dt} t^k = \sum_{n=0}^{\infty} \frac{1}{n!}\left( t \frac{d}{dt} \right)^n t^k = \sum_{n=0}^{\infty} \frac{k^n}{n!} t^k $$ since $t\frac{d}{dt}t^k = kt^k$. Thus $$ e^{td/dt} t^k = e^k t^k. $$ Ah, but this is $(et)^k$! Thus $$ e^{td/dt} f(t) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} e^{td/dt} t^k = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} (et)^k = f(et) $$ (assuming the expansion exists, of course). So while $e^{kd/dt}$ acts as a translation on the argument, $e^{td/dt}$ acts as a scaling: $$ e^{atd/dt} f(t) = f(e^a t). $$