At the foot of the mountain, the angle of elevation of its summit...

Let $P$ be the summit of the mountain and $Q$ be the foot.

Let $A$ be the first position and $B$ the second position of observation.

$BN$ and $BM$ are $⊥$s from $B$ to $PQ$ and $AQ$ respectively.
Then, $AB = 1000m = 1km$ ,
$∠MAB = {30}^{o},$
$∠MAP = {45}^{o},$

$∠NBP = {60}^{o}$
Now, $∠BAP = ∠MAP - ∠MAB = {45}^{o} - {30}^{o} = {15}^{o}$
$∠APB = ∠APN - ∠BPN = {45}^{o} - {30}^{o} = {15}^{o}$
$∴ ∆ ABP$ is isosceles

$∴ AB = BP$
But, $AB = 1$ kilometre

$∴ BP = 1$ kilometre.
Now, $PQ = PN + NQ = PN + BM$
$= BPsin {60}^{o} + ABsin {30}^{o}$
$= 1 ⋅ \dfrac{\sqrt{3}}{2} + 1 ⋅ \dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2} = 1 . 35k\text{m}$