Due to periodicity in $x$ you can write $f$ as a function on $(0,\infty)\times S^1$ so that your assumption in 1. becomes just uniform convergence on $S^1$ and this implies $f(t/\varepsilon,\cdot)\to a$ uniformly and hence $f(t/\varepsilon,x/\epsilon)\to a$. The same argument gives the second statement.

EDIT. Okay, here is a direct argument. $f(t,x)\to a$ (uniformly) on compact sets means that for all compact $K\subseteq \mathbb R$ and all $e>0$ there is $t_0>0$ such that for all $t>0$ and all $x\in K$ one has $|f(t,x)-a|<e$. Use this for $K=[0,1]$ together with the observation that $f(t/\varepsilon,x/\varepsilon^2)=f(t/\varepsilon,\lbrace x/\varepsilon^2\rbrace)$ where $\lbrace y\rbrace$ denotes the fractional part of a real number. This implies 1. since for $t>0$ it is enough to take $\varepsilon$ so small that $t/\varepsilon>t_0$. (If you want to have the convergence in 1 uniformly with respect to $t$, no chance.)

The proof of 2. is similar because every $x\in [0,1]$ can be expressed as $\lbrace y/\varepsilon^2\rbrace$ for some $y\in[0,1]$.