For x any number other than an integer, if x is between n and n+ 1, [x]= n so
x/[x]= x/n and x[x]= xn. It should be obvious that the function is continuous there. For x slightly below n, say x= n-y, x/[x]= n-y/(n-1) and x[x]= (n-y)n-1= xn/xy. For x slightly larger than n, say x= n+ y, x/[x]= (n+y)/n and x[x]= (n+y)n. Are the limits of those, as y goes to 0, the same?
As for [itex]sin(\pi x)[/itex] for x rational, 0 for x irrational, use the fact that [itex]\lim_{x\rightarrow a} f(x)= L[/itex] if and only if [itex]\lim_{n\rightarrow \infty} f(a_n)= L[/itex] for any sequence [itex]{a_n}[/itex] converging to x. In particular, for any number x, there exist a sequence of irrational numbers converging to x so for such a sequence the limit of this sequence will be 0. That means that in order that the limit itself exist at x, we must also have [itex]sin(a_n\pi)[/itex] converge to 0 also. Since sine is a continuous function, it is easy to see that that limit is [itex]sin(\pi x)[/itex]. That is, this function is continuous exactly for those x such that [itex]sin(\pi x)= 0[/itex].