All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X and Y denote the mass of a apple and pear in grams respectively.
$\mathbb{E}(X_1 + \ldots X_5) = 5(300) = 1500$

$\mathrm{Var}(X_1 + \ldots X_5) = 5(20^2) = 2000$

$X_1 + \ldots X_5 \sim \mathrm{N}(1500, 2000)$

$\mathrm{P}(X_1 + \ldots X_5 > 1600) \approx 0.0127$ (3 SF)

(ii)
$\mathbb{E}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(300) - 8(200) = -100$

$\mathrm{Var}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(20^2) + 8(15^2) = 3800$

$X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) \sim \mathrm{N}(-100, 3800)$

$\mathrm{P}(X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) > 0) \approx 0.0524$ (3 SF)

(iii)
Let A and B denote the mass of a peeled apple and peeled pear in grams respectively.
$\mathbb{E}(A)= \mathbb{E}(0.85X) = 0.85(300)=255$

$\mathbb{E}(B)= \mathbb{E}(0.90Y) = 0.90(200)=180$

$\mathrm{Var}(A) = \mathrm{Var}(0.85X) = 0.85^2(20^2) = 289$

$\mathrm{Var}(B) = \mathrm{Var}(0.90Y) = 0.90^2(15^2) = 182.25$

$\mathbb{E}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(255) + 8(180) = 2715$

$\mathrm{Var}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(289) + 8(182.25) = 2903$

$A_1 + \ldots A_5 + B_1 + \ldots B_8 \sim \mathrm{N}(2715, 2903)$

$\mathrm{P}(A_1 + \ldots A_5 + B_1 + \ldots B_8 ~ \textless ~ 2750) \approx 0.742$ (3 SF)