Write expressions w/out quantifiers (convert to AND/OR expressions) 

  
*A universe contains the three individuals $a,b$, and $c$. For these individuals, a predicate $Q(x,y)$ is defined, and its truth values are given by the following table:
  \begin{array}{c|ccc}
x\backslash y&a&b&c\\\hline
a&T&F&T\\
b&F&T&F\\
c&F&T&T
\end{array}
  Write each of the following expressions without quantifiers (i.e. covert them to expressions with ANDs and ORs or both) and then evaluate the expression.
  a) $\forall x\,\exists y\,Q(x,y)$
  b) $\forall y\,Q(y,b)$
  c) $\forall y\,Q(y,y)$
  

I found this post which says to convert all $\exists$ to OR and all $\forall$ to AND. However, my problem has 2 variables (instead of one, like in the aforementioned example). I'm not sure how to handle that.
Source.

My attempts:
a) $[P(a,a)\lor P(b,a)\lor P(c,a)]\land [P(b,a)\lor P(b,b)\lor P(b,c)]\land [P(c,a)\lor P(c,b)\lor P(c,c)]$
b) $P(a,b)\land P(b,b)\land P(c,b)$ 
c) $P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$
I am unsure how to evaluate the expressions..
To evaluate the expressions, I just need to match the "coordinates" with their respective T/F then evaluate whether or not the whole expression is T/F from there. Correct?
 A: 
$$\forall y ~ Q(y, y)$$
  $$P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$$

$$\forall y ~ Q(y, y)$$
Should be 
$$Q(a, a) \land Q(b, b) \land Q(c,c)$$
Otherwise you are correct.  Edit: As tunococ pointed out, in (a), $[P(a,a)\lor P(b,a)\lor P(c,a)]$ has a mistake not present in the other two terms.

To evaluate the expressions, I just need to match the "coordinates" with their respective T/F then evaluate whether or not the whole expression is T/F from there. Correct?

Yes.
A: 
However, my problem has 2 variables (instead of one, like in the aforementioned example). I'm not sure how to handle that.
a) $[P(a,a)\lor P(b,a)\lor P(c,a)]\land [P(b,a)\lor P(b,b)\lor P(b,c)]\land [P(c,a)\lor P(c,b)\lor P(c,c)]$

Almost as you did; but not quite (you muddled up the first grouping).  (Also, mind your Pees and Queues.)
When in doubt, write it out, just one step at a time. $$\begin{align}
&\qquad \forall x~\exists y~Q(x,y) 
\\ & \equiv \big(\exists y~Q(a, y)\big)\wedge\big(\exists y~Q(b, y)\big)\wedge \big(\exists y~Q(c, y)\big)
\\ & \equiv \big(\exists y~Q(a, y)\big)\wedge\big(\exists y~Q(b, y)\big)\wedge\big(Q(c,a)\vee Q(c,b)\vee Q(c,c)\big)
\\[1ex] & \equiv \big(Q(a,a)\vee Q(a,b)\vee Q(a,c)\big)\wedge\big(Q(b,a)\vee Q(b,b)\vee Q(b,c)\big)\wedge\big(Q(c,a)\vee Q(c,b)\vee Q(c,c)\big)
\\[1ex] & \equiv \big(\mathrm T\vee \mathrm F\vee \mathrm T\big)\wedge\big(\mathrm F\vee \mathrm T\vee \mathrm F\big)\wedge\big(\mathrm F\vee \mathrm T\vee \mathrm T\big)
\\ & \equiv (\mathrm T)\wedge(\mathrm T)\wedge(\mathrm T)
\\ & \equiv \mathrm T
\end{align}$$ 

b) $P(a,b)\land P(b,b)\land P(c,b)$ 

$\forall y\;Q(y,b) = Q(a,b)\land Q(b,b)\land Q(c,b) \quad\checkmark$ 

c) $P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$

$\forall y\;Q(y,y) = Q(a,a)\land Q(b,b)\land Q(c,c) \quad$ all $y$ within each term are replaced with the same instance.
