Invariance of canonical Hamiltonian equation when adding the total time derivative of a function of $q_i$ and $t$ to the Lagrangian The following is exercise 8.2 in 3rd edition (and exercise 8.19 in 2nd edition) of Goldstein's Classical Mechanics. 
Adding the total time derivative of a function of $q_i$ and t to the Lagrangian will not change the the Euler-Lagrangian equation. So if we make the following change to Lagrangian, 
$$L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q_1,q_2,...,q_n,t)}{dt}$$
we can get 
$$
\frac{d}{dt}\frac{\partial{L'}}{\partial{\dot{q_i}}} - \frac{\partial{L'}}{\partial{q_i}} = 0
$$
from 
$$
\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q_i}}} - \frac{\partial{L}}{\partial{q_i}} = 0
$$
How can we get the corresponding Hamiltonian equation part? This is to prove 
$$
\dot{p'_i} = \frac{\partial{H'}}{\partial q_i}
$$
$$
-\dot{q_i} = \frac{\partial{H'}}{\partial p'_i} 
$$
from 
$$
\dot{p_i} = \frac{\partial{H}}{\partial q_i}
$$
$$
-\dot{q_i} = \frac{\partial{H}}{\partial p_i} 
$$
where $p'_i = \frac{\partial L'}{\partial \dot q_i}$.
Edit
The corresponding $H'$ is
$$
H' = \sum_k{p'_k \dot{q_k}} - L'
$$
where $p'_k = \frac{\partial L'}{\partial \dot q_k}$.
 A: If $$ L \to L' = L +\frac{dF(q,t)}{dt}$$ the corresponding Hamiltonian becomes
$$ H \to H' = H - \frac{\partial F(q,t)}{\partial t} $$ as shown here. Moreover, the canonical momentum becomes $$ p \to P = p + \frac{\partial F}{\partial q}  $$ while $$ q \to Q = q $$ as shown here.
These formulas allow us to check the invariance of Hamilton's equations explicitly. Concretely, 
\begin{align}
\frac{dq}{dt}  &=  \frac{\partial H}{\partial p}  \notag \\ 
\end{align}
becomes
\begin{align}
 \frac{dQ}{dt}  &=  \frac{\partial H'}{\partial P} \\
\therefore \quad \frac{dq}{dt}  &=  \frac{\partial \left(H - \frac{\partial F(q,t)}{\partial t}  \right)}{\partial P} \\
\therefore \quad \frac{dq}{dt}  &= \frac{\partial H}{\partial P}  - \frac{\partial }{\partial P} \left( \frac{\partial F(q,t)}{\partial t}  \right) \\
\therefore \quad \frac{dq}{dt}  &= \frac{\partial H}{\partial P}    \\
\therefore \quad \frac{dq}{dt}  &= \frac{\partial H}{\partial p} \frac{\partial P}{\partial p}   \\
\therefore \quad \frac{dq}{dt}  &= \frac{\partial H}{\partial p} \quad \checkmark
\end{align}
where I used that $F$ does not depend on $P$ and $$\frac{\partial P}{\partial p} =\frac{\partial }{\partial p} \left( p+ \frac{\partial F}{\partial q}  \right) = 1. $$
Analogously, we can check Hamilton's second equation:
$$ \frac{dp}{dt}=  -\frac{\partial H(q,p,t)}{\partial q} .$$
However, there is a subtlety. After the transformation, we have on the right-hand side $\frac{\partial H'(Q,P,t)}{\partial Q}$. But here we need take into account that $p$ also depends on $q$, since $ p \to P = p + \frac{\partial F(Q,t)}{\partial Q}  $. Therefore
\begin{align}
\frac{\partial H'(Q,P,t)}{\partial Q} &= \frac{\partial H'(Q,p + \frac{\partial F}{\partial q} ,t)}{\partial Q} \\
&= \frac{\partial H'(Q,p,t)}{\partial Q} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \frac{\partial H(Q,p,t) }{\partial p} \frac{\partial p}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial \left(P- \frac{\partial F}{\partial q} 
 \right)}{\partial Q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{\partial^2 F(Q,t)}{\partial Q \partial t} - \dot Q \frac{\partial 
 }{\partial Q}  \frac{\partial F}{\partial q} \\
&= \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \,.
\end{align}
where we used that
$$ \frac{d}{dt} \frac{\partial F}{\partial Q}=  \frac{\partial^2 F(Q,t)}{\partial Q \partial t} + \dot Q \frac{\partial 
 }{\partial Q}  \frac{\partial F}{\partial q}  . $$
Using this, we can rewrite Hamilton's second equation after the transformation as follows:
\begin{align}
\frac{dP}{dt}&=  -\frac{\partial H'(Q,P,t)}{\partial Q} \\
\therefore \quad  \frac{d}{dt} \left( p+ \frac{\partial F(q,t)}{\partial q}  \right)  &=  \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q}   \\
\therefore \quad  \frac{dp}{dt} + \frac{d}{dt} \left(\frac{\partial F(q,t)}{\partial q}  \right)&=  \frac{\partial H(Q,p,t)}{\partial Q} - \frac{d}{dt} \frac{\partial F}{\partial Q} \\
\therefore \quad  \frac{dp}{dt} &= -\frac{\partial H}{\partial q} \quad \checkmark
\end{align}
EDIT: The subtlety was also noted here, but unfortunately without an answer and a few years ago there was even a paper which didn't notice it and claimed that Hamilton's equations are not invariant.
A: As you state in the comments,
$$
\frac{\mathrm dF}{\mathrm dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}
$$
So popping this into the Lagrangian,
$$
L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}
$$
The Hamiltonian $H=p\dot q-L$ implies
$$
H'=p'\dot{q}-L'=p\dot q+something\tag{1}
$$
where $something$ is for you to work out. Since $p=\partial L/\partial \dot q$, then we should assume that $p'=\partial L'/\partial\dot q$. It's not really necessary for this particular problem, but you can solve for $p'$.
The Hamiltonian formalism states that $q$, $\dot q$ and $p$ are independent, so we assume similarly that $q$, $\dot{q}$ and $p'$ are independent; hence $\partial L/\partial p=0\to\partial L'/\partial p'=0$. 
So now all you have to do is solve
$$
\frac{\partial H'}{\partial p'}\text{ and }-\frac{\partial H'}{\partial q}
$$
using Eq. (1) to see if the transformation in the Lagrangian preserves the Hamiltonian EOM (hint: it does). Note also that I assume a single coordinate $q$, there really isn't much of a difference between $q_i$ for $i=1$ and $i\in(1,N)$.
