Variance of a Sum of Brownian Values

Key idea. Write the sum in terms of independent increments.

Since $W_1+W_2 = 2W_1 + (W_2-W_1)$, and the increment $W_2-W_1$ is independent of $W_1$,

Here $W_1\sim\mathcal{N}(0,1)$ and $W_2-W_1\sim\mathcal{N}(0,1)$, so the variance is $4\cdot 1 + 1 = 5$.

Alternative. Use $\operatorname{Cov}(W_s,W_t)=\min(s,t)$:

In fact $W_1+W_2\sim\mathcal{N}(0,5)$, because it is a linear combination of jointly Gaussian variables.