Matrix function converges, how about the eigenvalues?

Parts (1) and (2) -- Yes. The coefficients of the characteristic polynomial are continuous functions of $A(t)$ (they are polynomials in the entries of $A(t)$!) and the roots of a polynomial are continuous functions of the coefficients.

Part (3): If $B$ has non-trivial Jordan blocks, this can fail. For example, $$\begin{pmatrix} 0 & 1 \\ -t & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + O(t)$$ as $t \to 0$, but the eigenvalues are $\pm \sqrt{t}$, which is not $O(t)$.

I believe that this is true when $B$ is diagonalizable. (I know that it is diagonalizable if $B$ has distinct eigenvalues, and I'll write up that argument if this one fails.)

We may as well assume $B$ is diagonalized. Let's concentrate on a particular eigenvalue $\lambda$ of $B$, with multiplicity $m$; we might as well assume $\lambda =0$. So $B = \mathrm{diag}(0,0,\cdots,0, \lambda_2, \lambda_3, ..., \lambda_k)$ with $m$ zeroes.

Suppose that $B-A(t)$ is $O(t^a)$. Explicitly expanding the determinant, the coefficient of $x^k$ in $\det(x \mathrm{Id} - B)$ is $O(t^{a(m-k)})$ for $k < m$, and the coefficient of $x^m$ does not go to $0$ as $t \to 0$. So the Newton polygon of the characteristic polynomial passes through $(m,0)$ and stays above the line from $(m,0)$ to $(0,ma)$. This shows that the bottom $m$ roots of the characteristic polynomial vanish at rate $O(t^a)$ or faster as $t \to 0$ (and the other roots do not vanish.)

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It occurs to me that it is worth sketching the argument for the Newton polygon claim directly so you can see how straightforward it is without learning the whole Newton polygon technology. Here is what I am claiming:

Lemma Fix $C>0$. Then there is a $D>0$ such that, if $f(u,z) = \sum_i f_i(u) z^i$ is holomorphic in $z$ on the disc of radius $1$, with $$|f_i(u)/f_m(u)| < \begin{cases} C u^{m-i} & i < m \\ C & i > m \end{cases}$$ then $f(u, \ )$ has $m$ roots in the disc of radius $D u$ for all sufficiently small $u$.

Proof sketch Consider $\oint \tfrac{f'}{f} dz$ where the integral is around the circle of radius $Du$. Write $f = f_m(u) z^m (1+\mbox{other terms})$ and $f' = m f_m(u) z^{m-1} (1+\mbox{other terms})$ so the integral is $\oint m \tfrac{dz}{z} + \mbox{other terms}$. Use the above conditions to bound the other terms.