For both parts (a) and (b), we have a clear upper bound of 2||f||1, by the triangle inequality. It remains to be shown that this is a tight upper bound in both cases.
The same approach can be used for both, with minor changes for the two different limits. Since step functions are dense in L1(R), pick epsilon e>0 and approximate f by some step function g, such that ||f-g||1<e. Let fx(t)=f(x+t), gx(t)=g(t+x).
Then ||fx+f|| = ||fx+f+(gx+g)-(gx+g)||. By the triangle inequality, this is greater than or equal to ||gx+g||-||fx+f-(gx+g)||. By yet another application of the triangle inequality, the second term is greater than or equal to -2e. The proof diverges at this point.
For part (a), for any particular t, x, |gx(t)+g(t)| is less than |gx(t)|+|g(t)| if and only if gx(t) and g(t) have opposite signs. Since g is a step function, this can clearly only happen when t and t+x are in intervals with different coefficients, and hence can happen at most for a distance of x per interval, across a finite number n of intervals. Since the difference for any particular step function g is bounded by M=max(g)-min(g), we get the following inequality:
||gx+g|| >= ||gx||+||g|| - x*n*M. Clearly, the limit of this as x goes to 0 is 2||g||, which is itself bounded below by 2||f||-2e. Adding up the two parts, we get a lower bound of the limit: limx->0||fx+f||>=2||f||-4e, for any positive epsilon. Thus the bound of 2||f|| is tight.
The justification in part (b) is simpler. Since g is a step function in L1, it has bounded support, so some value of x will be large enough that gx and g have disjoint supports. Hence we can just say that the limit is ||gx||+||g||>=2||f||-2e.