Let me restate what we want to prove:
$$\forall x\,\forall w \ \text{if $|w|$ is even, then } evens(w.x)=evens(w).evens(x)$$
For this, we fix an arbitrary $x$ and we prove the following by induction:
$$\forall w \ \text{if $|w|$ is even, then } evens(w.x)=evens(w).evens(x)$$
Now, one way to prove this is by induction on $|w|$:
If $|w|=0$, then $w=\epsilon$ and $|w|$ is even, so we must prove that $evens(\epsilon.x)=evens(\epsilon).evens(x)$. This is true because $evens(\epsilon)=\epsilon$ by definition.
Supposing the statement holds for $w$ up to a length of $k\geq 0$, consider a string $w$ of length $k+1$. We must show that
$$\text{if $k+1$ is even, then } evens(w.x)=evens(w).evens(x)$$
We distinguish two cases: $k+1$ is either odd or even.
If $k+1$ is odd, then the premise of the implication that we need to show is false. Therefore, the implication is true.
If $k+1$ is even, we need to show that $evens(w.x)=evens(w).evens(x)$. We write $w=a.b.t$, where $a,b$ are symbols and $|t|=k-1$. (We know we can do this because $k+1$ is even and $k+1\geq 1$, therefore $|w|=k+1\geq 2$.) By the definitions of the functions $evens$ and $odds$, we have:
$$evens(w.x)=evens(a.b.t.x)=odds(b.t.x)=b.evens(t.x)$$
But $|t|\leq k$, so we can use the inductive hypothesis to conclude that
$$evens(w.x)=b.evens(t.x)=b.evens(t).evens(x)$$
On the other hand, by the definitions of the functions $evens$ and $odds$:
$$evens(w).evens(x)=evens(a.b.t).evens(x)=odds(b.t).evens(x)=b.evens(t).evens(x)$$
Therefore, $evens(w.x)=evens(w).evens(x)$.
Since the OP was apparently going for a proof by induction on the structure of $w$ (i.e., assuming the statement holds for $t$, prove that it holds for $a.t$ where $a$ is a symbol), I'm including a proof along those lines. Unfortunately, in this case, the inductive hypothesis is not strong enough to allow us to conclude the proof. In cases like this, a standard technique is to strengthen the inductive hypothesis: we aim to prove a stronger statement than the one we originally wanted to prove, so that we can use a stronger inductive hypothesis during the proof. We will prove that
$$\begin{align}
\forall w\ & \text{if $|w|$ is even, then } evens(w.x) = evens(w) . evens(x) \ \wedge \\
& \text{if $|w|$ is odd, then } odds(w.x) = odds(w).evens(x)
\end{align}$$
If $w=\epsilon$, then $|w|=0$ is even, so we must prove that $evens(\epsilon.x)=evens(\epsilon).evens(x)$. This is true because $evens(\epsilon)=\epsilon$ by definition.
Supposing the statement holds for $t$, consider a string $w=a.t$, where $a$ is a symbol. We distinguish two cases: $|w|$ is either odd or even.
If $|w|$ is odd, then it suffices to show that $odds(w.x) = odds(w).evens(x)$. But $odds(w.x) = odds(a.t.x) = a.evens(t.x)$, and $|t|$ is even. Therefore, from the inductive hypothesis we obtain $evens(t.x)=evens(t).evens(x)$. We conclude that $odds(w.x)=a.evens(t).evens(x)$. On the other hand, $odds(w).evens(x)=odds(a.t).evens(x)=a.evens(t).evens(x)$, which proves the claim.
If $|w|$ is even, then it suffices to show that $evens(w.x)=evens(w).evens(x)$. But $evens(w.x)=evens(a.t.x)=odds(t.x)$, and $|t|$ is odd. Therefore, from the inductive hypothesis we obtain $odds(t.x)=odds(t).evens(x)$. We conclude that $evens(w.x)=odds(t).evens(x)$. On the other hand, $evens(w).evens(x)=evens(a.t).evens(x)=odds(t).evens(x)$, which proves the claim.
How does one guess the strengthened statement? It's a matter of understanding what the functions actually do. Here, $evens(w)$ just returns a string consisting of the symbols at even positions in $w$ (position numbering starts from $1$). Similarly, $odds(w)$ just returns a string consisting of the symbols at odd positions.
With this knowledge, it's actually not very surprising that $evens(w.x)=evens(w).evens(x)$ if $|w|$ is even: Sure, in order to write down the symbols at even positions of $w.x$, all you have to do is write down the symbols at even positions of $w$, all the way up to the last symbol of $w$ (which is at an even position because |w| is even), followed by the symbols at even positions of $x$.
Now, when you do the structural induction, you will have to deal with a string $t$ of odd length, so you need a similar property for odd-length strings. By a similar reasoning as in the previous paragraph, this property is $odds(w.x)=odds(w).evens(x)$.
