Part 1
I choose Christen Juzeszyn's function $ f(t)=5 \cos(2t) $.
One DT signal that could be generated from this would be $ f[n]=5 \cos(2n) $. This DT signal is non-periodic. $ \text{For }k,n \in \mathbb{N}, f[n+k] = 5 \cos(2(n+k)) = 5 \cos(2n+2k)). $ For f to be periodic, there needs to exist some natural number k (k > 0) such that f[n+k]=f[n]. i.e., $ 5 \cos(2n+2k) = 5 \cos(2n). $ For this to be true, we would need to find k such that for all n, $ 2n+2k = 2\pi (2n) \implies k=2\pi (n-1). $ However, $ 2\pi (n-1) \notin \mathbb{N} $ for any n, therefore no valid k exists.
Another DT signal that could be generated from this would be $ f[n]=5 \cos(2\pi n) $. This DT signal is periodic. For instance, let $ k=1 (k\in\mathbb{N}) $ be the period. Then, $ f[n+k]=f[n+1]=5 \cos(2\pi (n+1))=5 \cos(2\pi n+2\pi)=5 \cos(2\pi n)=f[n]\ \forall n\in\mathbb{Z}. $
Part 2
I chose Ben Horst's function $ x(t)=\frac{\sin(t)}{t} $. The function $ \sum_{k \in \mathbb{Z}} x(t+2 \pi k) = \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi k)}{t+2 \pi k} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k} $ is periodic.
Consider the period $ P = 2 \pi k $. $ x(t+P) |_{P=2 \pi} = \sum_{k \in \mathbb{Z}} \frac{\sin(t+P)}{(t+P)+2 \pi k}= \sum_{k \in \mathbb{Z}} \frac{\sin(t+2 \pi)}{(t+2 \pi)+2 \pi k}|_{P=2 \pi} $
$ = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi (k+1)} = \sum_{k \in \mathbb{Z}} \frac{\sin(t)}{t+2 \pi k} = x(t) \forall t \in \mathbb{R} $
