Q1. Is every sufficiently large composite integer a palindromic number in some non-trivial‡ base $B$? The trivial case of $n = 1(n-1) + 1 = (11)_{n-1}$ means it is palindromic in at least one base i.e., $n-1$. This MSE question and the sequence A016038 identifies non-palindromic numbers. In the notes for A016038, we have
All elements of the sequence greater than 6 are prime $(ab = a(b-1) + a$ or $a^2 = (a-1)^2 + 2(a-1) + 1)$. Mersenne and Fermat primes are not in the sequence.
Additional comments: if you can factor a number as $a*b$ then it is a palindrome in base $b-1$, where $b$ is the larger of the two factors. (If the number is a square, then it can be a palindrome in an additional way, in base $(\sqrt{n}-1))$. The $ab$ form does not work when $a = b-1$, but of course there are no two consecutive primes (other than 2,3, which explains the early special cases), so if you can factor a number as $a(a-1)$, then another factorization also exists).
‡: Non-trivial in this question means the base is not $n, n-1, b-1$ as defined in the quoted text above.
Q2: Is a palindromic number (in some base $B$, non-trivial again) easier to factor than a non-palindromic number? This linked article mentions the following about palindromic polynomials:
Any palindromic polynomial with real coefficients can be factored into a product of a constant times linear, quadratic, and quartic palindromic polynomials with real coefficients.
