Let $a,b,c$ be statements, $\Rightarrow$ is a tautological consequence.
Prove/disprove:
if $a, b\Rightarrow c$ then is it necessarily $a\Rightarrow c$ or $b\Rightarrow c$ ?
if $(a\Rightarrow c$ or $b\Rightarrow c)$ then is it necessarily $a, b\Rightarrow c$ ?
I think both are true,
since both $a,b$ are true then both $a\vee b$ are true and the tautology holds.
suppose WLOG $a$ is false, then we have a trivial implication with $a,b\Rightarrow c$
Recall that saying that a formula $\phi$ being a tautological consequence of set of formulas $\Gamma$ is, by definition, saying that any valuation that makes all formulas of $\Gamma$ true, also makes $\phi$ true. A couple of common abbreviations of this are $\Gamma \models \phi$ and $\Gamma \implies \phi$.
To address a comment made in the question, when $\Gamma$ is finite, i.e., when $\Gamma=\{\phi _1, \ldots ,\phi _n\}$ for some natural number $n$ and some formulas $\phi _1, \ldots ,\phi_n$, instead of $\{\phi _1, \ldots ,\phi _n\}\implies \phi$ it's common to write $\phi _1, \ldots ,\phi _n\implies \phi$.
Back to your question, let $a,b,c$ be formulas.
The first one doesn't necessarily hold. For an explicit counterexample consider two propositional atoms $p,q$ and define $a=p$, $b=q$ and $c=p\land q$ (with the appropriate conventions so that $c$ is actually a formula). Now prove that $a,b\implies c$ (I suggest you do this proving every detail) and note that neither $a\implies c$ (because of the valuation $v$ such that $v(p)=1$ and $v(q)=0$), nor $b\implies c$ hold (similar).
The second one holds and you can even relax the hypothesis. It holds that if $a\implies c$, then $a,b\implies c$. Assume that $a\implies c$, i.e., suppose that $$\text{for all valuations $v$, if $v(a)=1$, then $v(c)=1$}.$$ The goal is to prove that $$\text{for all valuations $v$, if $v(a)=1$ and $v(b)=1$, then $v(c)=1$}.$$
I'm hoping that me having written the problem more explicitly will be help enough for you to prove it.