Rom

$|5a-4|=a \Rightarrow (5a-4)=a \text{ or }4-5a=a\\ 5a-4=a\\ 4a=4\\ a=1\\ \\ 4-5a=a\\ 4=6a\\ a = \dfrac 2 3\\ 1 \cdot \dfrac 2 3 = \dfrac 2 3$

$-x + \dfrac 2 5 = -\dfrac 3 5 y\\ x = \dfrac 2 5 + \dfrac 3 5 y = \dfrac 1 5 (2+3y)\\ \text{now look at the second equation}\\ 3y = -\dfrac{18}{11}x + \dfrac{51}{11}\\ \text{substitute in the first result}\\ 3y = -\dfrac{18}{11}\left(\dfrac 1 5(2+3y)\right)+\dfrac{51}{11}$

$3y = -\dfrac{18}{55}(2+3y) + \dfrac{51}{11}\\ 3y+\dfrac{18\cdot 3}{55}y=-\dfrac{18\cdot 2}{55}+\dfrac{51}{11}\\ \dfrac{219}{55}y = \dfrac{219}{55}\\ y=1\\ x = \dfrac 1 5 (2+3y) = 1\\ (x,y) = (1,1)$

$10s = \dfrac{10}{3600} hr\\ d = \dfrac{10}{3600}hr \cdot 75 \dfrac{km}{hr} = \\ \dfrac{750}{3600} km = \dfrac{5}{24}km$

this is sort of cheating but the way I do it is recognizing that the coefficients are a line of Pascal's Triangle 

and thus they are binary coefficients.

That means that

$x^3 + 3x^2 + 3x + 1 = (x+1)^3$

I doubt they expect you to know that so you'll have to use the rational root theorem.

Given the coefficients of $x^3 \text{ and }1$ we know the only possible rational roots are $\pm 1$

$f(1) = 1 + 3 + 3 + 1 = 6 \neq 0\\ f(-1) = -1 + 3 -3 + 1 = 0 \\ \text{so -1 is a root}$

$\text{so }x+1 \text{ is a factor we then divide }\\ \dfrac{x^3 + 3x^2 + 3x + 1}{x+1} = x^2 + 2x + 1\\ \text{The result has the same possible two roots so we check as before}\\ \left . x^2 + 2x + 1 \right |_{x=1} = 4 \neq 0\\ \left . x^2 + 2x + 1 \right |_{x=-1} = 0\\ \text{again }(x+1) \text{ is a factor we again we divide}\\ \dfrac{x^2+2x+1}{x+1}=x+1 \\ \text{and we see all 3 factors are }x+1 \text{ i.e.}\\ f(x) = (x+1)^3$

And thus there is a single root, x=-1, that is repeated 3 times.

$p =\dfrac{52}{52}\dfrac{39}{52}\dfrac{26}{52}\dfrac{13}{52} = \dfrac{3}{32}$

I assume you were given the numbers in white.

I don't see any errors in (a)

she should expect (100+250)(0.2) = 70 balls.

$N_{left} = N_{girl}\cdot P[\text{girls are left handed}]+N_{boy}\cdot P[\text{boys are left handed}]\\ N_{left} = 320 \cdot 0.2 + 250 \cdot 0.1\\ N_{left} = 64+25 = 89$

$(a x + b)(b x + a) = ab x^2 + (a^2+b^2)x+ab = 26x^2 + c x + 26\\ \\ ab = 26\\ c = a^2 + b^2$

$\text{so basically we're minimizing }a^2+b^2 \text{ subject to }ab=26\\ \text{and subject to }a,b,c \in \mathbb{Z}$

$26 \text{ can be factored as }(2,13),~(1,26)\\ 2^2 + 13^2 = 173\\ 1^2 + 26^2 = 676\\ \text{and 173 is the smaller of those two so }\\ c=173$

$\text{conservation of momentum is a vector equation}\\ \text{assume the velocity of the bowling ball is }v_i = -2\hat{i} \\ \text{moving left along the x-axis if you like}\\ \text{when it recoils it's given to have the same speed}\\ \text{but now it's velocity will be }v_f = +2\hat{i}\\ \text{it's momentum of recoil is just }\\ m\vec{v} = 8kg \cdot (2\hat{i}~m/s) =16 \hat{i}~kg\cdot m/s\\ \text{the change is just twice this since it was originally exactly negative this}\\ \text{so }\Delta mv = 32\hat{i}~kg\cdot m/s$

$F = \dfrac{d~mv}{dt} \\ \text{and in this case this means}\\ F = \dfrac{\Delta mv}{t} \\ F = \dfrac{32\hat{i}}{0.5} = 64\hat{i} ~N\\ \text{note that }F \text{ is a vector. It has direction as expected}$

$\text{the key is to get them over a single denominator}\\ \dfrac 2 5 - \dfrac{3}{3y} = \\ \dfrac{2\cdot 3y - 3\cdot 5}{5\cdot 3y}=\\ \dfrac{6y-15}{15y} = \\ \dfrac{2y-5}{5y} =\\ \text{and you can go one step further if you like}\\ \dfrac 2 5 - \dfrac 1 y$