I'm having trouble deducing the value for the problem in the title.
Here is what I have done so far. (Given a standard Brownian motion (BM) $W_t, t\geq0 $ with $W_0 = 0$ and $\sigma^2=1$)
The standard covariance formula is given by $cov(W_x,W_y) = E(W_xW_y) - E(W_x)E(W_y)$ and because BM is also a Markov process $E(W_x)E(W_y) = 0$ so the covariance formula can be simplified to $cov(W_x,W_y)=E(W_xW_y)$
BM is also a Gaussian process so $cov(W_iW_j) = E(W_i,W_j) = min(W_i,W_j)$ where $i,j\in[0,\infty]$ so e.g. as $\sigma^2=1$, $ cov(W_7,W_9) = 7$
Going back to my original question of finding $cov(5W_7+6W_9,W_7)$, is the following computation correct?
$cov(5W_7+6W_9,W_7) = cov(5W_7,W_7) + cov(6W_9,W_7)$
$=E(5W_7,W_7) + E(6W_9,W_7)$
$=5E(W_7,W_7) + 6E(E_9,W_7)$
$=5*7 + 6*7$