In the context of propositional logic, a model is simply an assignment of truth values to the basic propositional variables. The statement $p\land q$ is true in any model that satisfies both $p$ and $q$, i.e., in which $p$ and $q$ are both true; the statement $p\lor q$, on the other hand, is true in any model that satisfies at least one of $p$ and $q$. This information is precisely what you’re collecting when you write out a truth table.
To see whether a formula $F_2$ is a logical consequence of a formula $F_1$, write out their joint truth table. Find the rows in which $F_1$ is true; those are essentially the models that satisfy $F_1$. Check to see whether $F_2$ is also true in all of those rows/models; if it is, then every model that satisfies $F_1$ automatically satisfies $F_2$, and $F_2$ is therefore a logical consequence of $F_1$. Here’s an example.
$$\begin{align*}
&F_1:\quad\big(p\to(q\lor r)\big)\land\lnot r\\
&F_2:\quad p\to q
\end{align*}$$
Here’s the combined truth table:
$$\begin{array}{c|c}
p&q&r&q\lor r&p\to(q\lor r)&\lnot r&p\to(q\lor r)\big)\land\lnot r&p\to q\\ \hline
T&T&T&T&T&F&F&\color{blue}{T}\\
\color{green}{T}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\
T&F&T&T&T&F&F&F\\
T&F&F&F&F&T&F&F\\
F&T&T&T&T&F&F&\color{blue}{T}\\
\color{green}{F}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\
F&F&T&T&T&F&F&\color{blue}{T}\\
\color{green}{F}&\color{green}{F}&\color{green}{F}&F&T&T&\color{red}{T}&\color{blue}{T}
\end{array}$$
The $T$’s for $F_1$ are in red, and those for $F_2$ are in blue. As you can see, in every row in which $F_1$ is true, $F_2$ is also true; the truth values of $p,q$, and $r$ in those rows are in green. $F_1$ holds in precisely those models in which $p$ and $q$ are true and $r$ is false, or $q$ is true and $p$ and $r$ are false, or all three are false. And in all such models $F_2$ holds as well, so $F_2$ is a logical consequence of $F_1$.
Note, though, that $F_2$ holds in some models in which $F_1$ does not hold, e.g., those in which $p,q$, and $r$ are all true. Thus, $F_1$ is not a logical consequence of $F_2$, and therefore $F_1$ and $F_2$ are not logically equivalent.
Why would you switch from "forall" to "exists" if you wanted to specify "$x$ is positive"?
You're going to want $$(\forall x > 0)(\exists a \exists b)(a^2+b^2 = x)$$
or, if your language doesn't let you formulate that, $$(\forall x)[x > 0 \to (\exists a \exists b)(a^2+b^2 = x)]$$
As an aside, the proposition is false: $3$ cannot be written as the sum of two squares. It is necessary and sufficient that primes $3 \pmod{4}$ appear only to even powers in the prime factorisation.
Best Answer
They are not equivalent. Yours is the converse of the desired statement. Translated into English, yours says:
Suppose that every student can access the internet; the statement in grey is still true — misleading, perhaps, but true. Your statement just lists two groups of people who can access the internet; it doesn’t say that these are the only groups who can do so.
The original statement, however, would not be true in this situation (assuming that there are some freshmen or some students not majoring in computer science): it says that the only people who can access the internet are comp. sci. majors and non-freshmen. And contrary to what you might at first think, it does not say that all comp. sci. majors and non-freshman can access the internet: it just says that if you’re not a comp. sci. major or a non-freshman, you definitely cannot access the internet.