The last point of your question kind of gives it away - in an arbitrage, the payoff is always $\geqslant 0$, regardless of the final state.
So, given what we have above:
$$
\begin{align}
S(t=0)&=100\\
S(t=1)&=100e^{5\%\ \cdot\ 1}&=105.13\\
S(t=1.5)&=100e^{5\%\ \cdot\ 1.5}&=107.79\\
\end{align}
$$
we will call the two options $C_1$ and $C_{1.5}$
So we can immediately see our options are both at the money, where $C_1$ is worth 11.92 and $C_{1.5}$ is worth 11.5. If we undiscount these prices, to get their values at expiry, we get 12.53 and 12.40 (respectively). Here is where we see the arbitrage - options at the same moneyness should be monotonically increasing in value as you increase the time to maturity.
Here we have the opposite $\rightarrow$ the shorter dated option is overpriced. When we're looking for an arbitrage, we short the overpriced and go long the underpriced, let's see what happens if we do that here:
Short 1 $C_1$, long $\frac{11.92}{11.5}$ $C_{1.5}$. Total cost = $+11.92 - \frac{11.92}{11.5} \cdot 11.5 = 0$. Now let's look at the payoff at a variety of states of the system:
After 1 year, there are two scenarios:
- $S(t=1) > 105.13$: $C_1$ expires costing you $S(t=1)-105.13$, but $C_{1.5}$ is worth at least $S(t=1)-105.13$ $\rightarrow$ portfolio value is positive.
- $S(t=1) \leqslant 105.13$: $C_1$ expires costing you nothing, and now you hold a 6m call option, and option values are always positive $\rightarrow$ portfolio value is positive.
So we see that at $t=1\mathrm{y}$, regardless of where the underlying ends up we have a positive portfolio value - this is arbitrage. You can see the equivalent option at $t=1\mathrm{y}$ to immediately realize the profit at that time (and invest it at the prevailing interest rate), or you can hold it to the end of the 1.5y period (at t=1y the value of doing either of these is the same).
the answer to these questions is very often the same - find out what is overpriced and short it to finance buying whatever is under priced.
