Ok, let's add some notation. Let $F$ mean that a NZer eats fruits and vegetables regularly, and $G$ mean that they consider a low-fat diet as a way of life. Add an $n$ in front to negate the letter. You are given $P(F) = 0.77$, $P(G) = 0.31$, and $P(G|F) = 0.35$. You should know that $P(A|B) = P(A\cap B)/P(B)$.

Part (a) is asking for $P(nG|F) = 1-P(G|F) = 1-0.35 = 0.65$

Part (b) is asking for $P(nG|nF)$. To figure this out, you need to figure out $P(nG \cap nF)$. From $P(nG|F) = 0.65 = P(nG \cap F)/P(F) = P(nG \cap F)/0.77$, you can solve for $P(nG \cap F) = 0.65\cdot 0.77 = 0.5005$. Now, $P(nG \cap F) + P(nG \cap nF) = P(nG)$. Plugging in the knowns, you have $P(nG \cap nF) = P(nG) - P(nG \cap F) = 1-P(G) - P(nG \cap F) = 1-0.31 - 0.5005 = 0.1895$. Finally, $P(nG|nF) = P(nG \cap nF)/P(nF) = 0.1895/(1-P(F)) = 0.1895/0.23 \approx 0.8239$

Part (c) is asking for $P(nF | G) = P(nF \cap G) / P(G)$. You are given that $P(G|F) = P(G \cap F) / P(F)$, so $P(F \cap G) = 0.35\cdot 0.77 = 0.2695$. Now, $P(F \cap G) + P(nF \cap G) = P(G) = 0.31$. So, $P(nF \cap G) = 0.31 - 0.2695 = 0.0405$. Finally, $P(nF | G) = 0.0405/0.31 \approx 0.1306$

Part (d) is asking for $P(nF | nG) = P(nF \cap nG) / P(nG)$. In part (b), we found $P(nF \cap nG) = 0.1895$, so this is $P(nF | nG) = 0.1895 / (1-P(G)) = 0.1895 / 0.69 \approx .2746$