Short answer: yes, there are numbers worse than $e$ and $\pi$. Almost all real numbers are worse. There are uncountably many reals, whereas any class of "nice transcendental numbers" you can think of will be only countable (and I propose exponential periods as a good class of numbers of the kind you describe).
The class of numbers which are roots of polynomials with rational coefficients are the algebraic numbers. The numbers which are definite integrals of algebraic functions with algebraic bounds are called the ring of periods. It is a larger class of numbers, including many familiar (suspected) transcendental numbers such as $\pi$, $\log 2$, $\zeta(3)$, and $\Gamma(p/q)^q$. See this nice primer by Kontsevich and Zagier.
Actually some common numbers like Euler's number $e$, the Euler-Mascheroni constant $\gamma$, and $1/\pi$ are still (suspected to be) missing in the ring of periods. So Kontsevich and Zagier go further and extend the ring to exponential periods, defined as integrals of products of exponentials of algebraic functions with algebraic functions. This gives a class of numbers that include "all algebraic powers of $e$, values of $\Gamma$ at rational arguments, values of Bessel functions, etc."
I assume that the exponential periods still don't include $\gamma$, because finally they claim that if you extend the class further by adding it, then you have "all classical constants in an appropriate sense" (whatever that means).
In this primer on exponential motives by Fresán and Jossen, according to Belkane and Brosnan, $\gamma$ is an exponential period, as witnessed by the integrals $\gamma=-\int_0^\infty\int_0^1e^x\frac{x-1}{(x-1)y+1}\,dy\,dx$ or $\gamma=-\int_0^\infty\int_1^x\frac{1}{y}e^{-x}\,dy\,dx$.
Anyway, the rational numbers are countable. The algebraic numbers are countable. The ring of periods is countable. The exponential periods are countable. And including $\gamma$ certainly still leaves you with a countable class of periods.
But the real numbers are uncountable, so it remains the case that most real numbers are not periods, not exponential periods, and cannot be written in any of these forms.
So to answer the question, yes, real numbers do get more exotic than numbers in the ring of periods or exponential periods like $e$ or $\pi$.
My understanding of statements of the form "$\Gamma$ takes values in the exponential periods at rational values" is that we would expect $\Gamma$ to take values at irrational arguments that are not exponential periods (although in general I expect proofs of such claims to be hard to come by).
So I would expect numbers like $\Gamma(\sqrt{2})$, $e^\pi$, $\zeta(\log 2)$ to be examples of computable numbers which are not exponential periods. But these would presumably not be considered “classical constants”.
Going further, even the computable numbers are countable, so most real numbers are not even computable, let alone in the ring of periods.
To see that $\alpha$ must be irrational, remember that rational numbers always have eventually periodic decimal (or indeed any base) expansions. But $\alpha$ clearly doesn't.
For question $2$, compare the denominators: for any $m$, when $k$ is sufficiently large we have $k!\cdot m<(k+1)!$, giving in turn $${1\over k!\cdot m}>{1\over (k+1)!}.$$ Now first imagine $c=c(\alpha)$, and then note that even if $c\not=c(\alpha)$ we still eventually get a contradiction as $k$ increases.
Best Answer
Take your favorite polynomial, $p(x) := x^2-x-1$, and your favorite transcendental number $c := \pi$. The number $p(c)$ is transcendental.
The key to transcendental numbers is polynomials with integer or rational coefficients. If $p(c)$ was algebraic, then there exists polynomials $q(x)$ such that $q(p(c))=0$ but then the composition $r(x):=q(p(x))$ is also a polynomial and $r(c)=0$ which would imply that $c$ is algebraic. The contrapositive statement is now proven.