We assume that OP is mainly concerned with condition (3').
A) Recall first that for forces ${\bf F}={\bf F}({\bf r})$ that do not dependent on velocity, we may use condition (3) to extend a definition of potential energy $U({\bf r}_{0})$ at a fiducial point ${\bf r}_{0}$ to a potential energy
$$\tag{A} U({\bf r}_{1})~:=~U({\bf r}_{0})-\int_{[0,1]} \!\mathrm{d}s~ {\bf F}({\bf r}(s)) \cdot {\bf r}^{\prime}(s) $$
at any other point ${\bf r}_{1}$ along any curve ${\bf r}: [0,1] \to \mathbb{R}^3$ with boundary conditions
$${\bf r}(s\!=\!0)={\bf r}_{0}\qquad\text{and}\qquad{\bf r}(s\!=\!1)={\bf r}_{1}.$$
Condition (3) ensures that definition (A) does not depend on the curve. Infinitesimally, definition (A) leads to
$$ \delta U~=~-{\bf F}\cdot \delta{\bf r}, $$
which implies condition (1).
A') We now repeat this for velocity dependent forces ${\bf F}={\bf F}({\bf r},{\bf v})$. The only difference is that points are replaced by paths that satisfy pertinent boundary conditions (BC). We may use condition (3') to extend a definition of potential action $S_{\rm pot}[{\bf r}_{0}]$ at a fiducial path ${\bf r}_{0}:[t_i,t_f]\to \mathbb{R}^3$ to a potential action
$$\tag{A'} S_{\rm pot}[{\bf r}_{1}]~:=~S_{\rm pot}[{\bf r}_{0}]-\int_{[t_i,t_f]\times [0,1]} \!\mathrm{d}t \wedge \mathrm{d}s~ {\bf F}({\bf r}(t,s),\dot{\bf r}(t,s)) \cdot {\bf r}^{\prime}(t,s) . $$
at any other path ${\bf r}_{1}:[t_i,t_f]\to \mathbb{R}^3$ along any homotopy ${\bf r}:[t_i,t_f]\times [0,1] \to \mathbb{R}^3$ with boundary conditions
$${\bf r}(t,s\!=\!0)={\bf r}_{0}(t)\qquad\text{and}\qquad{\bf r}(t,s\!=\!1)={\bf r}_{1}(t).$$
Condition (3') ensures that definition (A') does not depend on the homotopy. Infinitesimally, definition (A') leads to
$$ \delta S_{\rm pot}~=~-\int_{[t_i,t_f]} \!\mathrm{d}t~{\bf F} \cdot \delta{\bf r}, $$
which implies condition (1') under suitable BC.
The problem with the raised issue is that it is mainly formulated in the non-rigorous (but very effective) jargon of physicists. However, it can be re-formulated into a more mathematically precise version and the answer is straightforward.
First of all, if $A : D(A) \to H$ is selfadjoint, its spectum $\sigma(A)$ is the disjoint union of the continuous part of the spectrum $\sigma_c(A)$ and the point spectrum $\sigma_p(A)$. The latter is the set of eigenvectors.
Generally speaking, if $P: {\cal B}(\mathbb R) \to \mathfrak{B}(H)$ is a projection-valued measure defined on the Borel sets of the real line and $A,B$ are Borel set with $A \cap B =\emptyset$, we have $P_AP_B=0$ and
$$P_{A \cup B}= P_A + P_B\:.$$
Specializing to the case of $A= \sigma_p(A)$ and $B= \sigma_c(A)$ and $P=P^{(A)}$, the spectral measure of $A$, we conclude that:
$$P_{\sigma(A)}= P_{\sigma_p(A)} + P_{\sigma_c(A)}\:.$$
This identity propagates immediately to the spectral calculus giving rise to
$$A = \int_{\sigma_p(A)} \lambda dP^{(A)}(\lambda) + \int_{\sigma_c(A)} \lambda dP^{(A)}(\lambda)$$ where ($s-$ denoting the convergence in the strong topology)
$$\int_{\sigma_p(A)} \lambda dP^{(A)}(\lambda) = s-\sum_{u\in N} \lambda_u |u\rangle \langle u|$$ and $N$ is any Hilbert basis of the closed subspace $P^{(A)}_{\sigma_p(A)}(H)$ (viewed as a Hilbert space in its own right) made of eigenvectors $u$ with eigenvalues $\lambda_u$ (we may have $\lambda_u=\lambda_{u'}$ here).
It does not matter if $\lambda \in \sigma_p(A)$ is embedded in $\sigma_c(A)$, the above formulas are however valid. Everything is true when considering every Borel-measurable function $f$ of $A$:
$$f(A) = \int_{\sigma_p(A)} f(\lambda) dP^{(A)}(\lambda) + \int_{\sigma_c(A)} f(\lambda) dP^{(A)}(\lambda)$$ where
$$\int_{\sigma_p(A)} f(\lambda) dP^{(A)}(\lambda) = s-\sum_{u\in N} f(\lambda_u) |u\rangle \langle u|\:,$$
with the special case
$$I = \int_{\sigma_p(A)} dP^{(A)}(\lambda) + \int_{\sigma_c(A)} dP^{(A)}(\lambda)$$ where
$$\int_{\sigma_p(A)}dP^{(A)}(\lambda) = s-\sum_{u\in N} |u\rangle \langle u|$$
A mathematical remark helps understand why embedded eigenvalues do not create problems. The following facts are valid for $\lambda \in \mathbb{R}$.
(a) $\lambda \in \sigma_p(A)$ if and only if $P^{(A)}_{\{\lambda\}} \neq 0$,
(b) $\lambda \in \sigma_c(A)$ if and only if $P^{(A)}_{\{\lambda\}}= 0$, but
$P^{(A)}_{(\lambda-\delta, \lambda+\delta)}\neq 0$ for $\delta>0$
You see that the spectral measure $P^{(A)}$ cannot see the single points of $\sigma_c(A)$. And, if it sees a point, that point stays in $\sigma_p(A)$.
Furthermore, independently of the fact that some eigenvalues may be embedded in the continuous spectrum, $P^{(A)}_{\sigma_c(A)}(H)$ and
$P^{(A)}_{\sigma_p(A)}(H)$ are orthogonal subspaces of $H$.
ADDENDUM
A projector-valued measure (PVM) is a map $P : \Sigma(X) \to \mathfrak{B}(H)$ associating the sets $E$ of the $\sigma$-algebra $\Sigma(X)$ over $X$ to orthogonal projectors $P_E: H \to H$ on the Hilbert space $H$ such that
(1) $P(X)=I$,
(2) $P_EP_F = P_{E\cap F}$ if $E,F \in \Sigma(X)$,
(3) $P_{\cup_{j\in N} E_j}x = \sum_{n\in N}P_{E_n}x$ for every $x\in H$ and any family $\{E_n\}_{n\in N} \subset \Sigma(X)$ where $N \subset \mathbb{N}$ and $E_n \cap E_m = \emptyset$ if $n\neq m$.
(If $N$ is infinite, the sum refers to the topology of $H$.)
Given a PVM $P: \Sigma(X) \to \mathfrak{B}(H)$ and a pair of vectors $x,y\in H$, the map
$$\mu^{(P)}_{xy} : \Sigma(X) \ni E \mapsto \langle x| P_Ey\rangle \in \mathbb{C}$$
turns out to be a complex measure (with finite total variation). If $f:X \to \mathbb{C}$ is measurable, there exists an operator denoted by $$\int_X f dP : \Delta_f \to H$$ which is the unique operator such that
$$\left\langle x \left| \int_X f dP y \right. \right\rangle
= \int_X f(\lambda) d\mu^{(P)}_{xy}(\lambda)\quad x \in H\:, y \in \Delta_f\tag{1}$$
where the domain $\Delta_f$ is the dense linear subspace of $H$
$$\Delta_f := \left\{ y \in H \left| \int_X |f(\lambda)|^2 d\mu^{(P)}_{yy}(\lambda) < +\infty\right.\right\}\:.$$
With this definition, which is valid for every choice of $X$ and the $\sigma$-algebra $\Sigma(X)$, the spectral theorem for selfadjoint operators reads
THEOREM
If $A : D(A) \to H$ is a selfadjoint operator with domain $D(A) \subset H$, there exists a unique PVM $P^{(A)} : {\cal B}(\mathbb{R}) \to \mathfrak{B}(H)$, where ${\cal B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$, such that
$$\int_{\mathbb R} \lambda dP^{(A)}(\lambda) = A\:.$$
More precisely, the support of $P^{(A)}$ (i.e., the complement of the largest open set $O \subset \mathbb{R}$ with $P_O=0$) is exactly $\sigma(A)$.
Remarks
(1) With this result it is natural to define $$f(A) := \int_{\mathbb R} f(\lambda) dP^{(A)}(\lambda)=: \int_{\sigma(A)} f(\lambda) dP^{(A)}(\lambda)\:,$$
for every complex-valued Borel-measurable map defined on $\mathbb{R}$ (or more weakly on $\sigma(A)$).
(2)The above spectral theorem is still valid if replacing $A$ for a closed normal operator (in particular unitary operators). In that case $X= \mathbb{C}$. In the selfadjoint case $X = \mathbb{R}$ and, in this special case, the definition of a PVM can be given in terms of Stiltjes integrals, but this option is very limitative and for that reason it is rarely used nowadays.
(3) The notation $P(d\lambda)$ in place of $dP(\lambda)$ is a bit misleading (though the relevant object is the operator in the left hand side of Eq.(1) and not $dP$) since it refers to an apparently given measure $d\lambda$. But there is no such special choice in general. The standard measures arise once you give the vectors $x$ and $y$ and not before. In general there is no relation between $P^{(A)}$ and the Lebesgue measure $d\lambda$ on $\mathbb{R}$. Even if some forced (and sometime useful) relation can be found when comparing, e.g. $\langle x| P^{(A)}_E y \rangle$ and the Lebesgue measure $\int_E d\lambda$ and taking advantage of Lebesgue's theorem of decomposition of measures. This use leads to a further decomposition of the spectrum into absolutely continuous, singular continuous, and pure point parts. This decomposition, though useful in several contexts, is different from the original one into continuous, point, and residual which does not exploit the Lebesgue measure in any sense and that I used in my answer.
Unfortunately the sloppy language of physicists (I stress that I am a physicist!) does not permit to always appreciate some subtleties of these different decompositions and leads to misunderstandings and sometimes unmotivated issues. For instance the notions of discrete spectrum and point spectrum are different though they are often used as synonyms in the physical literature; the same happens for absolutely continuous spectrum and continuous spectrum.
Best Answer
You may be interested in the book "Methods of Applied Mathematics" by Arbogast and Bona, available here as a pdf
https://www.google.co.uk/url?q=https://www.ma.utexas.edu/users/arbogast/appMath08c.pdf&sa=U&ved=0ahUKEwjL5s2U-MPVAhWjLMAKHenGAAMQFggOMAA&usg=AFQjCNGW21X85j7nqyk6zuDOxRG5wdZ5Pg
Its quite a tough read, but of interest is proposition 1.39 on page 21, which can be restated for continuous functions that, if for all measurable $D \subset \Omega$, where $\Omega$ is the domain of our function (and a union of finite balls, so that we dont have any wierd measure zero bits), we have $$ \int_D f dV = 0 $$ Then $$ f(x) = 0 \forall x \in \Omega $$
This does you for case 1 when the right hand side is zero. I'm not at all sure that the nonzero case is even true...
For case 2 $$ \int_D f dV = \int_D g dV $$ $$ \int_D f - g dV = 0 $$ $$ f-g=0 $$ $$ f=g $$
I hope that answers part a). For part b) the answer is no. If you have a series of concentric spheres for which the volume integrals are equal, then you can then conclude that the surface integrals over the spheres are equal, but the values may be distributed differently over those spherical surfaces.
For part c), if those surfaces are not necessarily closed, then a version the above theorum will apply and you have that $F=G$ (I dont have a proof of this, just intuition). If the surfaces must be closed then the theorum will not apply since we do not have the result for all domains, but we can use the divergence theorum and then the above theorum to obtain that $\nabla \cdot F=\nabla \cdot G$
Edit
Here I correct and expand on my answer to part b).
For a simply connected open set $D$ and a continuous function $f$ we have the multivariate mean value theorem for integrals, which states that there exists some $c \in D$ for which $$ \int_D f(x) dV = f(c) |D| $$ where $|D|$ is the volume (specifically, the measure) of the open set.
Therefore, if we have an infinite sequence of simply connected open sets $D_0,D_1,...$ such that the sets all include some point $x_0$, and the sets 'tend towards' this point, specifically $$ \lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0 $$ then for each domain we can specify a point $c_n$ for which $$ \frac{1} {|D_n|} \int_{D_n} f(x) dV = f(c_n) =0 $$ where the equality with zero is from the question. We have that $c_n \in D_n$ thus $c_n \rightarrow x_0$ as $n \rightarrow \infty$ and $f(x_0)=0$.
Thus, if you find a set of simply connected open domains of arbitrarily small size around every point, the above shows that $f=0$.
We can therefore state the following theorem:
If, for each point $x_0$ in an open domain $\Omega$, we have that, for a sequence of simply connected open subsets $D_n$, $n$ any natural number, $$ \int_{D_n} f(x) dV =0 $$ for some continuous function $f$, and $$ \lim_{n \rightarrow \infty} (\max_{x \in D_n} |x-x_0|) = 0 $$ then $$ f=0 $$