Basic: Sequent definition, and-introduction, and iff I am reading through "Mathematical Logic by Ian Chiswell & Wilfred Hodges"(amazon, and publisher)
So far have it has covered $\land$-Introduction and $\land$-Elimination
Sadly this text only has answers to selected solutions, which annoys me to no end.

Exercise 2.3.3 (p.15) is
Show that $\{\phi_1, \phi_2\} \vdash \psi$ if and only if $\{(\phi_1 \land \phi_2)\} \vdash \psi$
I am a little confused as to what to show here.

I can state $\{\phi_1, \phi_2\} \vdash (\phi_1 \land \phi_2)$ as I can build this using $\land$-Introduction, but I don't follow how this involves if-and-only-if.
I think this is in part because I am a little confused as to what "$\{\phi_1, \phi_2\}$" means
- can this be read as "the set of assumptions we take to be true contains $\phi_1$ and $\phi_2$" ?
- or maybe "the set of assumptions we take to be true is exactly $\phi_1$ and $\phi_2$" ?

edit: for context I am reading through this for self-study, so I don't have the normal support of a classroom environment - and the lack of exercise solutions makes it hard to check my understanding.
 A: 
I don't follow how this involves if-and-only-if.

Hint
A) For: if $\{ (φ_1 ∧ φ_2) \} \vdash  ψ$, then $\{ φ_1, φ_2 \} \vdash ψ$.
1) $φ_1$ --- assumed
2) $φ_2$ --- assumed
3) $(φ_1 ∧ φ_2)$ --- from 1) and 2) by (∧I)
So far we have: 

$\{ φ_1, φ_2 \} \vdash (φ_1 ∧ φ_2)$.

Thus, by Sequent Rule (Transitive Rule) [page 8], from it and $\{ (φ_1 ∧ φ_2) \} \vdash  ψ$ we conclude with:

$\{ φ_1, φ_2 \} \vdash ψ$.


The Sequent Rule (Transitive Rule) says:

If $(Δ \vdash  ψ)$ is correct and for every $δ \in Δ$, $(Γ \vdash δ)$ is correct, then $(Γ \vdash ψ)$ is correct, where $Δ$ and $Γ$ are sets of formulae.

In your example, we have:

$Γ = \{ φ_1, φ_2 \}$ and $Δ = (φ_1 ∧ φ_2)$.

The derivation 1)-3) is $Γ \vdash δ$, where $(φ_1 ∧ φ_2)$ is the unique $δ \in Δ$, and the premise: $\{ (φ_1 ∧ φ_2) \} \vdash  ψ$ is $(Δ \vdash  ψ)$.
Thus, by transitivity, we conclude with: $(Γ \vdash ψ)$.


B) For: if $\{ φ_1, φ_2 \} \vdash ψ$, then $\{ (φ_1 ∧ φ_2) \} \vdash  ψ$, we have to use (∧E).
A: 
I think this is in part because I am a little confused as to what "{ϕ1,ϕ2}" means - can this be read as "the set of assumptions we take to be true contains ϕ1 and ϕ2" ? - or maybe "the set of assumptions we take to be true is exactly ϕ1 and ϕ2" ?

Indeed I think that's the only problem you're having here.
We write "$S \vdash φ$" to mean that we can derive the formula $φ$ by a sequence of rules from the formulae in the set $S$.
So for your exercise, the forward implication follows directly from $\land$-elimination, while the backward implication follows from $\land$-introduction.
If you need worked examples, you might want to try Stephen Simpson's notes.
