I need help understanding a question regarding a game played by two players. $A$ and $B$ play the following game: $A$ writes down
either number $1$ or number $2$, and $B$ must guess
which one. If the number that $A$ has written down
is $i$ and $B$ has guessed correctly, $B$ receives $i$ units
from $A$. If $B$ makes a wrong guess, $B$ pays $3/4$ unit to
$A$. If $B$ randomizes his decision by guessing $1$ with
probability $p$ and $2$ with probability $1 − p$, determine
his expected gain if 
$(a)$ $A$ has written down
number $1$.
$(b)$ $A$ has written down number $2$.
What value of $p$ maximizes the minimum possible
value of $B$’s expected gain, and what is
this maximin value? (Note that $B$’s expected
gain depends not only on $p$, but also on what
$A$ does.)
Consider now player $A$. Suppose that she also
randomizes her decision, writing down number $1$
with probability $q$. What is $A$’s expected loss if 
$(c)$ $B$ chooses number $1$?
$(d)$ $B$ chooses number $2$?
What value of $q$ minimizes $A$’s maximum
expected loss? Show that the minimum of $A$’s maximum
expected loss is equal to the maximum of $B$’s
minimum expected gain. 
My attempt (just the results)
$(a)$: $\frac{1}{4}(7p-3)$.
$(b)$: $2-\frac{11}{4}p$.
$(c)$: $\frac{1}{4}(3-7q)$.
$(d)$: $2-\frac{5}{4}q$.
I don't understand the what the "maximize minimum" and "minimize maximum" part of the question means. 
 A: Maximizing the minimum possible value of B's expected gain means that A either chooses 1 or 2, and you take the one that is less favorable to B (minimum expected gain). Then you choose the p for which this minimum will be of maximal value.
Minimax means that both players act perfectly rationally, so your opponent always makes the best move (min for you), and you need to choose your best move accordingly (max for you).
I think the last question is about finding the Nash equilibrium.
(I wanted to comment but I didn't have 50 reputation.)
A: The way that I understand "maximizing the minimum" is as follows:
Let $y(p)=\frac{7}{4}p - \frac{3}{4}$ and  $g(p)=2 - \frac{11}{4}p$. Let's plot the functions $y,g$ on a same graph and x axis being the value of $p \in [0,1]$. Because $y$ has positive slope and $g$ has negative slope, two lines most likely meet at a point $p \in [0,1]$. The meeting point is where the minimum of the two functions are maximized because, from that meeting point, if you select larger $p$, let's call it $p^*$ then $g(p^*)$ is going to be a new minimum that is smaller than we came from ($g(p) = y(p)$). Hope this helps! 
