If $A$ is false, is $\neg A$ true without invoking law of excluded middle? Let's say that we know that $A$ is false. We disallow the use of law of excluded middle. Then is it true that $\neg A$ is true?
Add: How would "false" be (usually) defined in intuitionisitc logic and classical logic? I think this can be a starting point for the question..
 A: The law of excluded middle is normally written as $A\vee\neg A$. In three-valued logic that would typically come out either true or "neither true nor false" but never false. If only always-true sentences count as theses, then excluded middle would not be a thesis--and yet $\neg A$ would be true if $A$ is false. This all assumes the most natural negation operator, namely one that flips true to false and false to true, and leaves "neither" as "neither."
A: The "usual" way to define the negation : $\lnot$ is to introduce a propositional constant (or $0$-connective) : $\bot$.
See Dirk van Dalen, Logic and Structure (5th ed - 2013), page 30 :

As usual “$\lnot \varphi$” is used here as an abbreviation for “$\varphi \rightarrow \bot$”.

In this way, in classical logic, the semantics for $\lnot$ is "reduced to" that for $\rightarrow$ :

when $\varphi$ is true $\varphi \rightarrow \bot$ is $T \rightarrow F$, which is false,

and 

when $\varphi$ is false $\varphi \rightarrow \bot$ is $F \rightarrow F$, which is true.

In intuitionistic logic the definition still holds.
What differ are the rules. In classical logic we have [see page 30] :

(RAA) $$\frac {\frac {[\lnot \varphi]} \bot } \varphi$$ 

which does not hold in the intuitionistic one.
The difference is due to the different semantics [see page 156] :

The primitive notion is here “$a$ proves $\varphi$”, where we understand by a proof a
  construction. [...] :
$(→)$ : $a$ proves $\varphi → \psi$ iff $a$ is a construction that converts any proof $p$ of $\varphi$ into a proof $a(p)$ of $\psi$.
$(⊥)$ : no $a$ proves $\bot$.

Thus [page 157] :

The only rule that lacks constructive content is that of reductio ad absurdum (RAA). As we have seen [see page 36], an application of RAA yields $\lnot \lnot \varphi → \varphi$, but for $\lnot \lnot \varphi → \varphi$ to hold [in intuitionistic logic] informally we need a construction that transforms a proof of $\lnot \lnot \varphi$ into a proof of $\varphi$. 
Now $a$ proves $\lnot \lnot \varphi$ if $a$ transforms each proof $b$ of $\lnot \varphi$ into a proof of $\bot$, i.e. there cannot be a proof $b$ of $\lnot \varphi$. $b$ itself should be a construction that transforms each proof $c$ of $\varphi$ into a proof of $\bot$. 
So we know that there cannot be a construction that turns a proof of $\varphi$ into a
  proof of $\bot$, but that is a long way from the required proof of $\varphi$ !

A: If A is false, the law of the excluded middle has no bearing on the status of ¬A.  The law of the excluded middle consists of a disjunction... as rather clearly seen when it gets written in Polish notation: ApNp.  The truth value of ¬A only depends on the unary operator ¬ (or N) here, as the following three-valued truth tables make clear:
A   0  1  2  N
0   0  1  2  1
1*  1  1  1  0
2   2  1  2  2

Here ApNp is not a tautology, since A2N2=A22=2.  However, if p=0 (p is false), then Np=1, in other words p is true.
