I will assume that "strict negativity of the real parts" means that there is a constant $c>0$ such that all eigenvalues of all matrices $A(t)$ have a real part less than $-c$.

(i) if $A(t)$ converges to a limit which is Hurwitz, i.e. for which all eigenvalues have negative real part, then you can use a quadratic Lyapunov function for that limit, which will also be a Lyapunov function for $A(t)$ for all $t$ sufficiently large. This is enough to ensure stability.

In detail: under said assumption on the limit A($\infty$) there exist positive definite matrices $P,Q$ such that

$A(\infty)^\top P + P A(\infty) = - Q $.

Now the Lyapunov function $V(x) = x^\top P x$ will satisfy

$\dot V (\varphi(t)) \leq - \tfrac{1}{2} \ \varphi(t)^\top Q \varphi(t)$

for all solutions $\varphi$ and all $t$ sufficiently large. Note that the factor $1/2$ is a smudge factor to handle the deviation of $A(t)$ from $A(\infty)$. Lyapunov's theorem now says that the zero position is asymptotically stable, in fact uniformly exponentially stable.

If you are looking for references to this and related facts look for the key words "slowly-varying systems".

(ii) This statement I cannot quite believe in its full generality. Maybe the authors have added an assumption somewhere that the matrix function $A(\cdot)$ is bounded? Otherwise please find a counterexample below. Let's first treat the case of bounded coefficients.

While the argument in (i) also works for complex matrices, the following is restricted to real matrices, as we will use monotonicity. If the entries of $A(t)$ have only finitely many maxima or minima up to $\infty$, this means that there is a time $T$ after which they all have no further extrema. This means that on the interval $[T,\infty)$ the entries are monotone. In which case they converge to a finite limit given that we assume that they are bounded.

If all entries converge we are back in the case (i), as then the limit matrix will be Hurwitz by continuous dependence of the eigenvalues of a matrix on the entries.

Now to the counterexample. Consider the system on $\mathbb{R}^2$ given by

$\dot{x} = \begin{bmatrix} -1 & e^t \\ 0 & -1 \end{bmatrix} x$

The eigenvalues are stuck at $-1$, the entries actually have no extrema at all, but the initial value problem for $x(0) = \begin{bmatrix} 1 & 1 \end{bmatrix}^\top$ has the solution

$\varphi(t) = \begin{bmatrix} 1 & e^{-t} \end{bmatrix}^\top$, $t\geq 0$.

So the zero position is not asymptotically stable - using linearity of the system.