I hope the below is what you asked for; it is based on the assumption that you asked to refute existence of conditionally, but not absolutely, convergent series $\sum a_n$ with the stated properties.
Unfortunately, both of your statements are incorrect.
Ad 1: Let $a_n = \dfrac{(-1)^n}n$. Then $\sum a_n$ converges conditionally; however it is clear that $\sum a_n - \dfrac1n$ does not converge at all.
Ad 2: Suppose $\sum a_n$ converges conditionally. Then since $\sum \dfrac1{n^2}$ converges absolutely, we have by the sum rule for limits (and hence, series) that $\sum a_n - \dfrac1{n^2}$ will converge conditionally as well.
Moreover, suppose $\sum a_n - \dfrac1{n^2}$ converged absolutely. Then by the triangle inequality:
$$|a_n| = \left|a_n - \frac1{n^2} +\frac1{n^2}\right| \le \left|a_n-\frac1{n^2}\right| + \frac1{n^2}$$
and since by assumption, the right-hand side is summable, so is the left-hand side; but this implies $\sum a_n$ converges absolutely.