In problem 3, the way you use the fact that the optimized part takes 1/6 of the total time after optimization is to realize that this means the unoptimized part must take 5/6 of the total time after optimization, or five times as long as the optimized part after optimization. Thus, if frac is the fraction of time spent on alu ops before the optimization, and speedup is the factor by which ALU ops are sped up, and 1.75 is the total speedup of the system, you can write Amdahl's law as an equation in two unknowns (frac and speedup). The unoptimized part takes (1-frac) and the optimized part takes frac/speedup after optimization, but the the fact noted in the first sentence of this paragraph tells you that

frac/speedup = (1/5)(1-frac)

so you can substitute (1/5)(1-frac) for frac/speedup in Amdahl's law, and now you have one equation in one unknown (frac) and you can solve for frac and then plug back in the value for frac to the original Amdahl's law equation and then solve for speedup.

3. Suppose a given optimization to the ALU speeds up execution for the system as a whole by a factor of 1.75. After optimization, ALU operations take up 1/6 of the total execution time. What fraction of execution time BEFORE OPTIMIZATION was taken up by ALU operations? What was the speedup factor to ALU operations due to the optimization